# Posts tagged as “LinkedList”

Given the `head` of a linked list, return the list after sorting it in ascending order.

Follow up: Can you sort the linked list in `O(n logn)` time and `O(1)` memory (i.e. constant space)?

Example 1:

```Input: head = [4,2,1,3]
Output: [1,2,3,4]
```

Example 2:

```Input: head = [-1,5,3,4,0]
Output: [-1,0,3,4,5]
```

Example 3:

```Input: head = []
Output: []
```

Constraints:

• The number of nodes in the list is in the range `[0, 5 * 104]`.
• `-105 <= Node.val <= 105`

Idea:

Merge Sort (use slow and fast pointer to find the middle of the linked list and split them)

Time Complexity: O(nlogn)

Solution:

``````/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var sortList = function(head) {
// empty or one node
if (head === null || head.next === null)
let mid = getMid(head);
let left = sortList(head);
let right = sortList(mid);
return merge(left, right);
};

function merge(list1, list2) {
let newHead = new ListNode();
let tail = new ListNode();
while (list1 !== null && list2 !== null) {
if (list1.val < list2.val) {
tail.next = list1;
list1 = list1.next;
tail = tail.next;
} else {
tail.next = list2;
list2 = list2.next;
tail = tail.next;
}
}
tail.next = (list1 !== null) ? list1 : list2;
};

let slow = head;
let fast = head;
let midHead = null;
while (fast !== null && fast.next !== null) {
slow = slow.next
fast = fast.next.next;
}
let secondHalf = midHead.next;
// !!!Important here:
// disconnect, split them into two lists
return secondHalf;
}``````

Given the `head` of a singly linked list, reverse the list, and return the reversed list.

Example 1:

```Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
```

Example 2:

```Input: head = [1,2]
Output: [2,1]
```

Example 3:

```Input: head = []
Output: []
```

Constraints:

• The number of nodes in the list is the range `[0, 5000]`.
• `-5000 <= Node.val <= 5000`

Idea: 背口诀

```reverse LinkedList 口诀: create newHead and next

Solution:

``````/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function(head) {
// empty or only one node
if (head === null || head.next === null) return head;

let newHead = new ListNode();
let next = new ListNode();
while (head !== null) {
// next helper create
// disconnect, and point to newHead.next
// move head to next location
}
};``````

You are given the `root` of a binary search tree (BST) and an integer `val`.

Find the node in the BST that the node’s value equals `val` and return the subtree rooted with that node. If such a node does not exist, return `null`.

Example 1:

```Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
```

Example 2:

```Input: root = [4,2,7,1,3], val = 5
Output: []
```

Constraints:

• The number of nodes in the tree is in the range `[1, 5000]`.
• `1 <= Node.val <= 107`
• `root` is a binary search tree.
• `1 <= val <= 107`

Solution:

``````/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} val
* @return {TreeNode}
*/
var searchBST = function(root, val) {
if (root === null) return null
if (root.val === val)
return root;
else if (val < root.val)
return searchBST(root.left, val);
else
return searchBST(root.right, val);
};``````

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

Example 1:

```Input: head = [1,2,3,4]
Output: [2,1,4,3]
```

Example 2:

```Input: head = []
Output: []
```

Example 3:

```Input: head = [1]
Output: [1]
```

Constraints:

• The number of nodes in the list is in the range `[0, 100]`.
• `0 <= Node.val <= 100`

Solution:

``````/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var swapPairs = function(head) {
if (head === null) return null;
let dummy = head;
while (head !== null && head.next !== null) {
if (head.next !== null)
}
return dummy;
};

}``````

Given the `head` of a linked list, remove the `nth` node from the end of the list and return its head.

Follow up: Could you do this in one pass?

Example 1:

```Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
```

Example 2:

```Input: head = [1], n = 1
Output: []
```

Example 3:

```Input: head = [1,2], n = 1
Output: [1]
```

Constraints:

• The number of nodes in the list is `sz`.
• `1 <= sz <= 30`
• `0 <= Node.val <= 100`
• `1 <= n <= sz`
``````/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @param {number} n
* @return {ListNode}
*/
var removeNthFromEnd = function(head, n) {
if (head.next === null)
return null;

let first = new ListNode();
let second = first;
// first point to n + 1
first = first.next;
while (n > 0) {
first = first.next;
n--;
}
// first already reach to the end null, means need to remove the first node
if (first === null) {
} else {
// maintaining N nodes apart between first and second pointer
while (first !== null) {
first = first.next;
second = second.next;
}
// Now second pointer to the node that need to be removed
// remove that node now
second.next = second.next.next;
}

};``````

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example 1:

Input: l1 = [2,4,3], l2 = [5,6,4]

Output: [7,0,8]

Explanation: 342 + 465 = 807.

Example 2:

Input: l1 = [0], l2 = [0]

Output: [0]

Example 3:

Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]

Output: [8,9,9,9,0,0,0,1]

Constraints:

• The number of nodes in each linked list is in the range `[1, 100]`.
• `0 <= Node.val <= 9`
• It is guaranteed that the list represents a number that does not have leading zeros.
```/**
* Definition for singly-linked list.
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {
let head = new ListNode();
let temp = head;
let sum = 0;
let carry = 0;
while (l1 !== null || l2 !== null) {
sum = carry + (l1 !== null ? l1.val : 0) + (l2 !== null ? l2.val : 0);
carry = sum >= 10 ? 1 : 0;
sum = sum % 10;

temp.next = new ListNode(sum);
temp = temp.next;

if (l1) l1 = l1.next;
if (l2) l2 = l2.next;

if (carry) temp.next = new ListNode(carry);
}