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Posts published in “Linked List Operations”

LeetCode 148. Sort List (javascript)

By stone on April 15, 2021

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) 
        return head;
    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();
    tail = newHead;
    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;
    return newHead.next;
};

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

LeetCode 206. Reverse Linked List

By stone on April 11, 2021

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
next -> head.next -> newHead.next -> head -> 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
        next = head.next;
        // disconnect, and point to newHead.next
        head.next = newHead.next;
        // move newHead.next to head
        newHead.next = head;
        // move head to next location
        head = next;
    }
    return newHead.next;
};

LeetCode 700. Search in a Binary Search Tree (javascript)

By stone on April 9, 2021

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);
};

LeetCode 24. Swap Nodes in Pairs (javascript)

By stone on April 9, 2021

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;
    if (head.next === null) return head;
    let dummy = head;
    while (head !== null && head.next !== null) {
        swap(head);
        if (head.next !== null)
            head = head.next.next;
    }
    return dummy;
};

function swap(head) {
    [head.val, head.next.val] = [head.next.val, head.val];
    return head;
}

LeetCode 202. Happy Number (javascript)

By stone on March 27, 2021

Write an algorithm to determine if a number n is happy.

A happy number is a number defined by the following process:

  • Starting with any positive integer, replace the number by the sum of the squares of its digits.
  • Repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1.
  • Those numbers for which this process ends in 1 are happy.

Return true if n is a happy number, and false if not.

Example 1:

Input: n = 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

Example 2:

Input: n = 2
Output: false

Constraints:

  • 1 <= n <= 231 - 1

Idea:

create a function to calculate the sum of the squares of its digits.

use fast and slow pointers traversal concept (the other one execute function twice) to check if they can reach 1 (means it is happy number)

Solution:

/**
 * @param {number} n
 * @return {boolean}
 */
var isHappy = function(n) {
    let slow = n;
    let fast = n;
    
    do {
        slow = squareSum(slow);
        fast = squareSum(squareSum(fast));
    } while (slow !== fast);
    
    return fast === 1;
};

var squareSum = function(n) {
    let sum = 0;
    while(n) {
        sum += (n % 10) * (n % 10);
        n = parseInt(n / 10);
    }
    return sum;
};

LeetCode 19. Remove Nth Node From End of List (Javascript)

By stone on March 19, 2021

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();
    first.next = head;
    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) {
        head = head.next;
    } 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;
    }

    return head;
    
};