Linked List

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Java Variables & Data Types

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Java Variable & Data Types

Primitive Type
- int a = 5;
- byte, short, int, long, float, double, char, boolean
- ONLY these eight types.
Reference Type
- String str = new String("this is a string");
- String, int[], MyObject, ..., etc.
- new & array.

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Memory

Heap Memory

Stack Memory

Java Variable & Data Types

Primitive Type - Stack Memory
Reference Type - Heap Memory

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Variable Define - Primitive

Primitive Type
- Store the value directly.
Reference Type
- Store the address of the real object.

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Variable Define - Primitive

int a = 5

Memory

Heap Memory

Stack Memory

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Variable Define - Primitive

int a = 5
Store the value 5 directly in the stack memory

Memory

Heap Memory

Stack Memory

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Variable Define - Primitive

int a = 5
Store the value 5 directly in the stack memory
Give 5 an alias, 'a'

Memory

Heap Memory

Stack Memory

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Variable Define - Reference

MyObject obj = new MyObject();

Memory

Heap Memory

Stack Memory

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Variable Define - Reference

MyObject obj;
obj = new MyObject();

Memory

Heap Memory

Stack Memory

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Variable Define - Reference

MyObject obj;
obj = new MyObject();

Memory

Heap Memory

Stack Memory

obj

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Variable Define - Reference

MyObject obj;
obj = new MyObject();

Memory

Heap Memory

Stack Memory

MyObject instance

obj

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Variable Define - Reference

MyObject obj;
obj = new MyObject();

Memory

Heap Memory

Stack Memory

MyObject instance

obj

addr1

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Variable Define - Reference

MyObject obj = new MyObject();
Date date = new Date(1, 1, 2016);

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj

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Variable Define - Reference

MyObject obj = new MyObject();
Date date = new Date(1, 1, 2016);

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj

1
1
2016

Date instance

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Variable Define - Reference

MyObject obj = new MyObject();
Date date = new Date(1, 1, 2016);

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj

1
1
2016

addr2

Date instance

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Variable Define - Reference

MyObject obj = new MyObject();
Date date = new Date(1, 1, 2016);

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj

1
1
2016

addr2

date

Date instance

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Variable Assignment - Primitive

Primitive Type
- Assign the value of RHS to the LHS.
Reference Type
- Assign the address of RHS object to the LHS.

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Variable Assignment - Primitive

int a = 5;
int b = a; // a = 5, b = 5

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Variable Assignment - Primitive

int a = 5;
int b = a;
b = 4; // a = 5, b = 4

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Variable Assignment - Primitive

int a = 5;
int b = a;
b = 4; // a = 5, b = 4
No mater how you assign a with b, or assign a/b with another value
- changing one of them, will NOT affect the other one.

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Variable Assignment - Reference

MyObject obj1 = new MyObject();

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj1

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Variable Assignment - Reference

MyObject obj1 = new MyObject();
MyObject obj2 = obj1; // obj1 and obj2 points to the same object instance.

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj1

obj2

addr1

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Variable Assignment - Reference

MyObject obj1 = new MyObject();
MyObject obj2 = obj1; // obj1 and obj2 point to the same object instance. Changing the field of one will affect the other
// Suppose obj1 has a field 'a', its value is 5
obj2.a = 4; // obj1.a = 4

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Variable Assignment - Reference

MyObject obj1 = new MyObject();
MyObject obj2 = obj1;
obj2 = new MyObject();

Memory

Heap Memory

Stack Memory

MyObject instance

addr1

obj1

obj2

addr2

MyObject instance

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Variable Assignment - Reference

MyObject obj1 = new MyObject();
MyObject obj2 = obj1;
obj2 = new MyObject();
// obj1 and obj2 point to different objects. Changing one will not affect the other.
// Suppose obj1 has field 'a', and its value is 5.
obj2.a = 3; // obj1.a = 5

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Variable Comparison

Primitive Type
- Compare the value of the variable.
- int a = 5;
- int b = 5; // a == b ? True
Reference Type
- Compare the address of the object.
- MyObject obj1 = new MyObject();
- MyObject obj2 = obj1; // obj1 == obj2 ? True
- obj2 = new MyObject(); // obj1 == obj2 ? False

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Variable Comparison

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String a = "abc";
String b = "abc";
System.out.println(a == b); // true
String c = new String("abc");
System.out.println(a == c); // false
System.out.println(a.equals(c)); // trye
String d = new String("abc");
System.out.println(c == d); // false
System.out.println(d.equals(c)); // true

Integer num1 = 1;
Integer num2 = 1;
System.out.println(num1 == num2); // true
Integer num3 = new Integer(1);
System.out.println(num1 == num3); // false
Integer num4 = new Integer(1);
System.out.println(num3 == num4); // false
System.out.println(num3.equals(num4)); // true

// OPTIONAL OPTIONAL******
Integer num5 = new Integer(0);
System.out.println(num3 == num4+num5); // true

Use equals function for reference comparison

Passing Parameters

Primitive Type
- Passing the COPY of primitive value.
Reference Type
- Passing the COPY of object address.

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Passing Parameters

Primitive Type
- Passing the copy of primitive value.

public void increase(int a) {
    a = a + 1;
    System.out.print(a);
}

public static void main() {
    int a = 5;
    increase(a);
    System.out.print(a);
}

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Passing Parameters

Primitive Type
- Passing the copy of primitive value.

public void increase(int a) {
    a = a + 1;
    System.out.print(a);
}

public static void main() {
    int a = 5;
    increase(a); // print 6.
    System.out.print(a); // print 5
}

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Passing Parameters

Primitive Type
- Passing the copy of primitive value.
- The original variable will NEVER be changed.

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void increase(MyObject obj) {
    obj.a += 1;
    System.out.print(obj.a);
}

public static void main() {
    MyObject obj = new MyObject();
    obj.a = 5;
    increase(obj);
    System.out.print(obj.a);
}

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void increase(MyObject obj) {
    obj.a += 1;
    System.out.print(obj.a);
}

public static void main() {
    MyObject obj = new MyObject();
    obj.a = 5;
    increase(obj); // print 6.
    System.out.print(obj.a); // print 6
}

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Passing Parameters

Reference Type
- Passing the copy of object address.
- Changing the object (parameter) inside the function WILL change the original object.

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void increase(MyObject obj) {
    MyObject obj2 = obj;
    obj2.a += 1;
    System.out.print(obj.a);
    System.out.print(obj2.a);
}

public static void main() {
    MyObject obj = new MyObject();
    obj.a = 5;
    increase(obj);
    System.out.print(obj.a);
}

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void change(String str) {
    str = new String("world");
    System.out.print(str);
}

public static void main() {
    String s = new String("hello");
    change(s); // print 'world';
    System.out.print(s); // print 'hello'.
}

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void increase(MyObject obj) {
    MyObject obj2 = obj;
    obj2.a += 1;
    System.out.print(obj.a);
    System.out.print(obj2.a);
}

public static void main() {
    MyObject obj = new MyObject();
    obj.a = 5;
    increase(obj); // print 6, then print 6.
    System.out.print(obj.a);
}

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Passing Parameters

Reference Type
- Passing the copy of object address.

public void increase(MyObject obj) {
    MyObject obj2 = obj;
    obj2.a += 1;
    System.out.print(obj.a);
    System.out.print(obj2.a);
}

public static void main() {
    MyObject obj = new MyObject();
    obj.a = 5;
    increase(obj); // print 6, then print 6.
    System.out.print(obj.a); // print 6.
}

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Java Data Type Summary

Primitive Type
- Always means a literal value, such as 5, 's', true, etc.
Reference Type
- Always means an address.
- ... = new Object();
- ... = obj;
- myFunction(obj);

Address of the real object.

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Linked List

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What is Linked List?

A data structure consisting of a linear collection of data elements, called nodes pointing to the next node by means of pointer.

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1	4	8	2	5	3

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1	4	8	2	5	3

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1	4	8	2	5	3

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Linked List

Each node knows the address of its next node.
The first node can represent the whole linked list.
Accessing any element in the linked list, needs going through all nodes before it. O(n).
Sometimes, the node also knows the address of its previous node, then the list is known as "Doubly Linked List".

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Linked List in Code

public class ListNode {

int val;

ListNode next;

ListNode(int val_) {

val = val_;

next = null; // this line is optional

}

public class LinkedList {

ListNode head;

ListNode tail;

int size;

}

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WHY use linked list?

For most cases, the size of memory needed CANNOT be known before.

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Basic Operations

Get: <index> ---> value, O(n)
Set: <index, value> ---> void (new LinkedList), O(n)
Add: <opt_index, value> ---> void (new LinkedList), O(n)
Remove: <index/value> ---> void (new LinkedList), O(n)

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How to implement LinkedList

public class LinkedList{
    // TODO: implement this class.
    public int get(int index);
    public void set(int index, int value);
    public void add(int index, int value);
    public void remove(int index);
    public void removeByValue(int value);
}

public class ListNode {
    int val;
    ListNode next;
    public ListNode(int val_) {
        this.val = val_;
    }
}

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How to implement LinkedList - Fields

public class LinkedList{
    
    private ListNode head;
    private ListNode tail;
    private int size;

    public LinkedList() {
        head = null;
        tail = null;
        size = 0;    
    }
}

Java will automatically initialize the value for the fields.

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How to implement LinkedList - Fields

public class LinkedList{
    
    private ListNode head;
    private ListNode tail;
    private int size;
    
    public LinkedList() {
    }
}

Java will automatically generate a non-param constructor if you don't have other constructors.

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How to implement LinkedList - Fields

public class LinkedList{
    
    private ListNode head;
    private ListNode tail;
    private int size;
 
}

Java will automatically generate a non-param constructor if you don't have other constructors.

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How to implement LinkedList - Fields

public class LinkedList{
    
    private ListNode head = null;
    private ListNode tail = null;
    private int size = 0;
 
}

Recommend !!!

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How to implement LinkedList - Get

public class LinkedList{
    public void checkBoundsExclusive(int index) {
        if (index < 0 || index >= size) {
            // throw Exception.
        }
    }
        
    public ListNode getEntry(int index) {
        ListNode cur = head;
        for (int i = 0; i < index; i++) {
            cur = cur.next;    
        }
        return cur;
    }
    
    public int get(int index) {
        checkBoundsExclusive(index);
        return getEntry(index).val;
    }
}

public ListNode getEntry(int index) {
    ListNode cur = head;
    while (index-- != 0) {
        cur = cur.next;    
    }
    return cur;
}

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How to implement LinkedList - Set

public class LinkedList{
    
    public void set(int index, int value) {
        checkBoundsExclusive(index);
        Listnode node = getEntry(index);
        node.val = value;    
    }
 
}

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode newNode = new ListNode(value);
        ListNode pre = getEntry(index-1);
        newNode.next = pre.next;
        pre.next = newNode;
    }
 
}

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode newNode = new ListNode(value);
        if (index == 0) {
            newNode.next = head;
            head = newNode;
            return;
        }
        ListNode pre = getEntry(index-1);
        newNode.next = pre.next;
        pre.next = newNode;
    }
 
}

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How to implement LinkedList - Remove

public class LinkedList{
    
    public void remove(int index) {
        checkBoundsExclusive(index);
        ListNode pre = getEntry(index-1);
        pre.next = pre.next.next;
    }
 
}

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How to implement LinkedList - Remove

public class LinkedList{
    
    public void remove(int index) {
        checkBoundsExclusive(index);
        if (index == 0) {
            head = head.next;
            return;
        }
        ListNode pre = getEntry(index-1);
        pre.next = pre.next.next;
    }
 
}

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Why need special case

Because head node is different

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null

There is no node pointing at head!!!

head

Head Node is Different

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null

There is no node pointing at head!!!

?

head

Head Node is Different

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null

head

fake head

Head Node is Different

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null

head

dummy

dummy.next can represent original linked list.
head node will have a previous node.
head node is NOT different any more.

Head Node is Different

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        
        ListNode pre = dummy;
        while (index-- != 0) {
            pre = pre.next;
        }        
        ListNode newNode = new ListNode(value);
        newNode.next = pre.next;
        pre.next = newNode;
        head = dummy.next;
    }
 
}

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        
        ListNode pre = dummy;
        while (index-- != 0) {
            pre = pre.next;
        }        
        ListNode newNode = new ListNode(value);
        newNode.next = pre.next;
        pre.next = newNode;
        head = dummy.next;
    }
 
}

ListNode pre = getEntry(index-1);

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        
        ListNode pre = dummy;
        while (index--) {
            pre = pre.next;
        }        
        ListNode newNode = new ListNode(value);
        newNode.next = pre.next;
        pre.next = newNode;
        head = dummy.next;
    }
 
}

ListNode pre = head;
index = index - 1;
while (index-- != 0) {
    pre = pre.next;
}

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How to implement LinkedList - Add

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        while (index-- != 0) {
            pre = pre.next;
        }        
        ListNode newNode = new ListNode(value);
        newNode.next = pre.next;
        pre.next = newNode;
        head = dummy.next;
    }
 
}

public class LinkedList{
    
    public void add(int index, int value) {
        checkBoundsExclusive(index);
        ListNode newNode = new ListNode(value);
        if (index == 0) {
            newNode.next = head;
            head = newNode;
            return;
        }
        ListNode pre = head;
        index = index - 1;
        while (index-- != 0) {
            pre = pre.next;
        }
        newNode.next = pre.next;
        pre.next = newNode;
    }
 
}

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Dummy Node

Make head node not different any more.
Simplify edge case situation.
- Make code SHORTER.
  - shorter is better.
- FEWER mistakes.
There will always exist AT LEAST ONE node, but the linked list starts from dummy's next.

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How to implement LinkedList - Remove

public class LinkedList{
    
    public void remove(int index) {
        checkBoundsExclusive(index);
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode pre = dummy;
        while (index-- != 0) {
            pre = pre.next;
        }
        pre.next = pre.next.next;
        head = dummy.next;
    }
 
}

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Summary for Linked List

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Summary for Linked List

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Summary for Linked List

Remove

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If data structure of the linkedlist will be changed, dummy node is needed.
- add, remove.
One node can only be accessed by its previous node.
- Don't change the previous code's next before this node is assigned to a new node, unless it's useless.

Summary for Linked List

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Linked List in interview

For most cases, there will only one head node given.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        
    }
}

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Linked List in interview

For most cases, there will only one head node given.
- No size.
- Check null when using while.
  - while (cur != null) { // do something }
  - while (pre.next != null) { // do something }
Communication before you write code !!!
- size, index range, tail, etc.

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Exercises

public class ListNode {

int val;

ListNode next;

public ListNode(int val) {

this.val = val;

}

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Linked list length.
Kth node from the end.
Mid node of the list.
Whether circle exists.

Count Related

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Linked List Length

public int length(ListNode head) {
    ListNode cur = head;
    int length = 0;
    while (cur != null) {
        length++;
        cur = cur.next;
    }
    return length;
}

Given a linked list, return the length of it.

Examples:

Input: 1->4->2->3

Output: 4

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Kth node from the end

Given a linked list, return the kth node from the end. Linked list will never be empty and k will always be valid.

Examples:

Input: 1->4->2->3, 2

Output: 2

Input: 3->5->9->6->8, 3

Output: 9

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public ListNode kthNodeFromEnd(
    ListNode head, int k) {

    int length = length(head);
    int index = length - k;
    ListNode cur = head;
    int count = 0;
    while (count < index) {
        cur = cur.next;
        count++;
    }
    return cur;
}

Kth node from the end

public ListNode kthNodeFromEnd(
    ListNode head, int k) {
    
    int length = length(head);
    int index = length - k;
    ListNode cur = head;
    while (index-- != 0) {
        cur = cur.next;
    }
    return cur;
}

Given a linked list, return the kth node from the end. Linked list will never be empty and k will always be valid.

Examples:

Input: 1->4->2->3, 2

Output: 2

Input: 3->5->9->6->8, 3

Output: 9

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Kth node from the end

public ListNode kthNodeFromEnd(
    ListNode head, int k) {
    
    ListNode first = head;
    while (k-- != 0) {
        first = first.next;
    }
    ListNode second = head;
    while (first != null) {
        first = first.next;
        second = second.next;
    }
    return second;
}

Given a linked list, return the kth node from the end. Linked list will never be empty and k will always be valid.

Examples:

Input: 1->4->2->3, 2

Output: 2

Input: 3->5->9->6->8, 3

Output: 9

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Middle Node

public ListNode midNode(ListNode head) {
    int length = length(head);
    int index = (length - 1) / 2;
    ListNode cur = head;
    while (index-- != 0) {
        cur = cur.next;
    }
    return cur;
}

Given a linked list, return the middle node. Linked list will never be empty.

Examples:

Input: 1->4->2->3

Output: 4

Input: 3->5->9->6->8

Output: 9

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Middle Node

public ListNode midNode(ListNode head) {
    ListNode fast = head;
    ListNode slow = head;
    while (fast.next != null &&
           fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    return slow;
}

Given a linked list, return the middle node. Linked list will never be empty.

Examples:

Input: 1->4->2->3

Output: 4

Input: 3->5->9->6->8

Output: 9

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Linked List Cycle

public boolean hasCycle(ListNode head) {
    if (head == null) {
        return false;
    }
    ListNode fast = head;
    ListNode slow = head;
    while (fast != null) {
        if (fast.next == null) {
            return false;
        }
        if (fast.next == slow) {
            return true;
        }
        fast = fast.next.next;
        slow = slow.next;
    }
    return false;
}

Given a linked list, define if there is a cycle in it.

Examples:

Input: 1->4->2->3

Output: false.

Input: 3->5->9->3(original)

Output: true

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Linked List Cycle - Follow up

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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2 *(k+l1) = k + l1 + l2 + l1

=> k = l2

Linked List Cycle - Follow up

public ListNode detectCycle(ListNode head) {
    ListNode fast = head, slow = head;
    while (fast != null && slow != null) {
        if (fast.next != null) {
            fast = fast.next.next;
        } else {
            return null;
        }
        slow = slow.next;
        if (fast == slow) {
            ListNode temp = head;
            while (temp != slow) {
                temp = temp.next;
                slow = slow.next;
            }
            return slow;
        }
    }
    return null;
}

Given a linked list, return the node where the cycle begins or null.

Examples:

Input: 1->4->2->3

Output: null.

Input: 3->5->9->3(original)

Output: 3

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Exercise Summary 1.

head == null?
k ? compare(k, length) && compare(k, 0) && range(k): nothing.
Null Pointer Exception
- while (node != null)
  - node.next
  - ~~node.next.next~~
Test Cases

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Structure Related

Remove Duplicate from Sorted List
Merge Sorted List
Reverse Linked List
Swap Nodes in Pairs

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Remove Duplicate

public ListNode removeDuplicates(
    ListNode head) {
    
    if (head == null) {
        return null;
    }
    ListNode pre = head;
    while (pre.next != null) {
        if (pre.val == pre.next.val) {
            pre.next = pre.next.next;
        } else {
            pre = pre.next;
        }
    }
    return head;
}

Given a sorted linked list, remove all duplicates such that each element appear only once.

Examples:

Input: 1->2->2->2->3->3

Output: 1->2->3

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Remove Duplicate - Follow up

public ListNode removeDuplicates(ListNode head) {
    if (head == null) {
        return null;
    }
    ListNode dummy = new ListNode(-1);
    dummy.next = head;
    ListNode pre = dummy;

    while (pre.next != null && pre.next.next != null) {
        if (pre.next.val == pre.next.next.val) {
            int lastVal = pre.next.val;
            while (pre.next != null && pre.next.val == lastVal) {
                pre.next = pre.next.next;
            }
        } else {
            pre = pre.next;
        }
    }
    
    return dummy.next;
}

Given a sorted linked list, remove all nodes that have duplicate numbers leaving only the distinct numbers from the original list.

Examples:

Input: 1->2->2->2->3->3

Output: 1

Input: 3->4->4->5->6

Output: 3->5->6

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Merge Sorted List

Merge two sorted list and return it as a new list.

Examples:

Input: 1->2->2->2, -1->3->4->5

Output: -1->1->2->2->2->3->4->5

public ListNode merge(ListNode head1, ListNode head2) {
    ListNode dummy = new ListNode(-1);
    ListNode cur = dummy;
    while (head1 != null && head2 != null) {
        if (head1.val < head2.val) {
            cur.next = new ListNode(head1.val);
            head1 = head1.next;
        } else {
            cur.next = new ListNode(head2.val);
            head2 = head2.next;
        }
        cur = cur.next;
    }
    while (head2 != null) {
        cur.next = new ListNode(head2.val);
        head2 = head2.next;
        cur = cur.next; 
    }
    while (head1 != null) {
        cur.next = new ListNode(head1.val);
        head1 = head1.next;
        cur = cur.next; 
    }
    return dummy.next;
}

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Merge Sorted List

Merge two sorted list and return it as a new list.

Examples:

Input: 1->2->2->2, -1->3->4->5

Output: -1->1->2->2->2->3->4->5

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public ListNode merge(ListNode head1, ListNode head2) {
    ListNode dummy = new ListNode(-1);
    ListNode cur = dummy;
    while (head1 != null && head2 != null) {
        if (head1.val < head2.val) {
            cur.next = head1;
            head1 = head1.next;
        } else {
            cur.next = head2;
            head2 = head2.next;
        }
        cur = cur.next;
    }
    if (head1 == null) {
        cur.next = head2;
    } else if (head2 == null) {
        cur.next = head1;
    }
    return dummy.next;
}

Reverse Linked List

Reverse a given linked list.

Examples:

Input: 1->2->3->4

Output: 4->3->2->1

public ListNode reverse(ListNode head) {
    ListNode pre = null;
    ListNode cur = head;
    while (cur != null) {
        ListNode temp = cur.next;
        cur.next = pre;
        pre = cur;
        cur = temp;
    }
    return pre;
}

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Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

Examples:

Input: 1->2->3->4

Output: 2->1->4->3

Note: Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

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Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.

Examples:

Input: 1->2->3->4

Output: 2->1->4->3

Note: Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

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public ListNode swapPairs(ListNode head) {
    ListNode dummy = new ListNode(-1);
    dummy.next = head;
    ListNode pre = dummy;
    while (pre.next != null && pre.next.next != null) {
        ListNode first = pre.next, second = pre.next.next;
        first.next = second.next;
        second.next = first;
        pre.next = second;
        pre = first;
    }
    return dummy.next;
}

Exercise Summary 2.

dummy.
- Structure (including head) will be changed.
Null Pointer Exception
- while (node != null)
  - node.next
- while (pre.next != null)
  - pre.next, pre.next.next
Test Cases

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Exercise Summary 2.

dummy.
- Structure (including head) will be changed.
Null Pointer Exception
- while (node != null)
  - node.next
- while (pre.next != null)
  - pre.next, pre.next.next
Test Cases

 while (pre.next != null && pre.next.next != null) {
    if (pre.next.val == pre.next.next.val) {
        int lastVal = pre.next.val;
        while (pre.next != null &&
               pre.next.val == lastVal) {
            pre.next = pre.next.next;
        }
    } else {
        pre = pre.next;
    }
}

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Delete Node in a Linked List

Given one node in a linked list, remove that node. (Only the reference to the specific node is given).

Examples:

Input: 1->2->3->4, 3

Output: 1->2->4

public void removeNode(ListNode node) {
    if (node == null) {
        return;
    }
    if (node.next == null) {
        node = null;
        return;
    }
    node.val = node.next.val;
    node.next = node.next.next;
    return;
}

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Summary

Dummy node
Don't lose nodes
- Store the latter part before modifying the current one.
Practice

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Homework

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Homework (Optional)

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Given two linked list (without circle), detect if they intersect with each other.

Implement:

Java: public boolean detectIntersection(ListNode head1, ListNode head2);

c++: bool detectIntersection(ListNode* head1, ListNode* head2);

Follow up: Return null or the intersection ListNode.

Java: public ListNode detectIntersection(ListNode head1, ListNode head2);

c++: ListNode* detectIntersection(ListNode* head1, ListNode* head2);

Follow up 2: What if the linked lists may have circle? Return null or the intersection node.

Homework (Optional)

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