We read every piece of feedback, and take your input very seriously.
To see all available qualifiers, see our documentation.
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
Difficulty: 简单
Related Topics: 递归, 链表
将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1:
输入:l1 = [1,2,4], l2 = [1,3,4] 输出:[1,1,2,3,4,4]
示例 2:
输入:l1 = [], l2 = [] 输出:[]
示例 3:
输入:l1 = [], l2 = [0] 输出:[0]
提示:
[0, 50]
-100 <= Node.val <= 100
l1
l2
Language: JavaScript
/** * Definition for singly-linked list. * function ListNode(val, next) { * this.val = (val===undefined ? 0 : val) * this.next = (next===undefined ? null : next) * } */ /** * @param {ListNode} list1 * @param {ListNode} list2 * @return {ListNode} */ // 递归 // var mergeTwoLists = function(list1, list2) { // if (!list1) { // return list2 // } else if (!list2) { // return list1 // } else if (list1.val < list2.val) { // list1.next = mergeTwoLists(list1.next, list2) // return list1 // } else { // list2.next = mergeTwoLists(list1, list2.next) // return list2 // } // } // 迭代 var mergeTwoLists = function(list1, list2) { const dummy = new ListNode() let curr = dummy while (list1 && list2) { list1.val < list2.val ? [curr.next, list1] = [list1, list1.next] : [curr.next, list2] = [list2, list2.next] curr = curr.next } curr.next = list1 ?? list2 return dummy.next }
The text was updated successfully, but these errors were encountered:
No branches or pull requests
21. 合并两个有序链表
Description
Difficulty: 简单
Related Topics: 递归, 链表
将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1:
示例 2:
示例 3:
提示:
[0, 50]
-100 <= Node.val <= 100
l1
和l2
均按 非递减顺序 排列Solution
Language: JavaScript
The text was updated successfully, but these errors were encountered: