LeetCode每日一练【21】

Merge Two Sorted Lists

我的解法

我的第一次提交

介绍

.....

思路

1. 构建链表: listNode(存放返回结果), listnode1(链表1), listnode2(链表2)

2. 创建三个指针分别指向三个链表: let [pointer, pointer1, pointer2] = [listnode, listnode1.next, listnode2.next];

3. 依次比较listnode1和listnode2的val值

   1. 如果pointer1.val < pointer2.val, 移动pointer1指针, pointer1.val值入pointer指向的链表, 移动pointer指针
   2. 如果pointer1.val > pointer2.val, 移动pointer2指针, pointer2.val值入pointer指向的链表, 移动pointer指针

4. 如果pointer1和pointer2在经过循环比较后, 可能存在一个非空, 需要将pointer.next = poninter1 或者 pointer.next = pointer2

5. 返回listnode.next

代码

/*
 * @Author: fox
 * @Date: 2022-05-03 10:48:54
 * @LastEditors: fox
 * @LastEditTime: 2022-05-03 13:50:45
 * @Description: https://leetcode.com/problems/merge-two-sorted-lists/
 */

class ListNode {
    constructor (val, next) {
        this.val = (val === undefined ? 0 : val);
        this.next = (next === undefined ? null : next)
    }
}

/**
 * @description: Runtime: 40.85%   Memory Usage: 83.57%
 * @param {ListNode} list1
 * @param {ListNode} list2
 * @return {ListNode}
 */
const mergeTwoLists = (list1, list2) => {
    const listnode = new ListNode(0);
    const listnode1 = new ListNode(0, list1);
    const listnode2 = new ListNode(0, list2);

    let [pointer, pointer1, pointer2] = [listnode, listnode1.next, listnode2.next];

    while (pointer1 && pointer2) {
        if (pointer1.val < pointer2.val) {
            pointer.next = new ListNode(pointer1.val)
            pointer1 = pointer1.next
            pointer = pointer.next
        } else {
            pointer.next = new ListNode(pointer2.val)
            pointer2 = pointer2.next
            pointer = pointer.next
        }
    }

    if (pointer1) {
        pointer.next = pointer1
    }

    if (pointer2) {
        pointer.next = pointer2
    }

    return listnode.next
};

const list1 = [1, 2, 4]
const list2 = [1, 3, 4]

console.log(mergeTwoLists(list1, list2))