### Find Operation

It takes us `O(N)` time on average to `visit an element by index`, where N is the length of the linked list.

#### Add a Node after a Given Node

If we want to add a new value after a given node `prev`, we should:

Initialize a new node `cur` with the given value;

Link the "next" field of `cur` to prev's next node `next`;

Link the "next" field in `prev` to `cur`.

O(1) time, space complexity

#### Add a Node at the Beginning

So it is essential to update`head`when adding a new node at the beginning of the list.

1. Initialize a new node `cur`;
2. Link the new node to our original head node `head`.
3. Assign `cur` to `head`.

### Doubly Linked List Node Structure

``````// Definition for doubly-linked list.
class DoublyListNode {
int val;
DoublyListNode next, prev;
DoublyListNode(int x) {val = x;}
}
``````

## Linked List Basic Operations and Classic Problems

Based on Java, Singly Linked List

### ListNode Class

``````class ListNode {
int val;
ListNode next;

ListNode(int x) { val = x; }
}
``````

iterative

``````public ListNode reverseList(ListNode head) {
ListNode prev = null;
}
return prev;
}
``````
``````public ListNode reverseList(ListNode head) {
ListNode prev = null;
while (curr != null) {
ListNode nextTemp = curr.next;
curr.next = prev;
prev = curr;
curr = nextTemp;
}
return prev;
}
``````

recursive

``````public ListNode reverseList(ListNode head) {
return p;
}
``````
``````    ListNode newHead;
return null;
}
}
ListNode reverseUtil(ListNode curr, ListNode prev) {

/* If last node mark it head*/
if (curr.next == null) {

/* Update next to prev node */
curr.next = prev;

}

/* Save curr->next node for recursive call */
ListNode next = curr.next;

/* and update next ..*/
curr.next = prev;

reverseUtil(next, curr);
}
``````

### Find the Middle Point

If there are two middle nodes, return the second middle node.

``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
while (fast != null && fast.next != null) {
slow = slow.next;
fast = fast.next.next;
}
return slow;
}
}
``````

If there are two middle nodes, return the first middle node.

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

### Find the Nth Element

``````ListNode fast = head;
for (int i = 0; i < n - 1; i++) {
fast = fast.next;
if (fast == null) {
return null;
}
}
``````

### Dummy Node

``````ListNode dummy = new ListNode(Integer.MIN_VALUE);
ListNode current = dummy;
...
while (current != null) {
...
current.next = blahblahblah;
...
current = current.next;
}
...
return dummy.next;
``````

### Merge Two Sorted Lists

Iterative

``````public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode lastNode = dummy;

while (l1 != null && l2 != null) {
if (l1.val < l2.val) {
lastNode.next = l1;
l1 = l1.next;
} else {
lastNode.next = l2;
l2 = l2.next;
}
lastNode = lastNode.next;
}

if (l1 != null) {
lastNode.next = l1;
} else {
lastNode.next = l2;
}

return dummy.next;
}
``````
``````/**
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode p = dummy;
while (l1 != null || l2 != null) {
if (l1 == null) {
p.next = l2;
break;
}
if (l2 == null) {
p.next = l1;
break;
}
if (l1.val < l2.val) {
p.next = l1;
l1 = l1.next;
} else {
p.next = l2;
l2 = l2.next;
}
p = p.next;
}
return dummy.next;
}
}
``````

Recursive

``````public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
} else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
``````

Remove all elements from a linked list of integers that have value val.

Iterative

``````public ListNode removeElements(ListNode head, int val) {
ListNode dummy = new ListNode(0);
ListNode prev = dummy;
while (curr != null) {
if (curr.val == val) {
prev.next = curr.next;
} else {
prev = prev.next;
}
curr = curr.next;
}
return dummy.next;
}
``````

Recursive

``````public ListNode removeElements(ListNode head, int val) {
if (head == null) return null;
}
``````

``````public Boolean hasCycle(ListNode head) {
return false;
}

ListNode fast, slow;
while (fast != slow) {
if(fast==null || fast.next==null)
return false;
fast = fast.next.next;
slow = slow.next;
}
return true;
}
``````

## Two Pointer Technique in Linked List 链表中的两个指针

For Linked List, we can use two-pointer technique:

Two pointers are `moved at different speed`: one is faster while another one might be slower.

The scenario, which is also called `slow-pointer and fast-pointer technique`, is really useful.

`If there is no cycle, the fast pointer will stop at the end of the linked list.`

`If there is a cycle, the fast pointer will eventually meet with the slow pointer.`

The proper speed for the two pointers?

Slow pointer - One step at a time

Fast pointer - Two steps at a time

#### 链表中快慢指针的Template

``````// Initialize slow & fast pointers
/**
* Change this condition to fit specific problem.
* Attention: remember to avoid null-pointer error
**/
while (slow != null && fast != null && fast.next != null) {
slow = slow.next;           // move slow pointer one step each time
fast = fast.next.next;      // move fast pointer two steps each time
if (slow == fast) {         // change this condition to fit specific problem
return true;
}
}
return false;   // change return value to fit specific problem
``````
##### Tips

1. Always examine if the node is null before you call the next field.

Getting the next node of a null node will cause the null-pointer error. For example, before we run `fast = fast.next.next`, we need to examine both `fast` and `fast.next` is not null.

2. Carefully define the end conditions of your loop.

Run several examples to make sure your end conditions will not result in an endless loop. And you have to take our first tip into consideration when you define your end conditions.

### Performance Comparison of ArrayList vs LinkedList

Reverse List in Java

https://www.techiedelight.com/reverse-arraylist-java/

https://www.geeksforgeeks.org/collections-reverse-java-examples/

## Data Structure Time Complexity Comparison

Here we provide a comparison of`time complexity`between the linked list and other data structures includingarray, queue and stack:

After this comparison, it is not difficult to come up with our conclusion:

If you need to add or delete a node frequently, a linked list could be a good choice.

If you need to access an element by index often, an array might be a better choice than a linked list.