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Reverse a linked list from position m to n. Do it in-place and in one-pass.
For example:
Given 1->2->3->4->5->NULL
, m = 2 and n = 4,
return 1->4->3->2->5->NULL
.
Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.
https://leetcode.com/problems/reverse-linked-list-ii/
翻转从m到n的链表。
翻转的操作还是和上一题一样:http://www.cnblogs.com/Liok3187/p/4540490.html
1 /** 2 * Definition for singly-linked list. 3 * function ListNode(val) { 4 * this.val = val; 5 * this.next = null; 6 * } 7 */ 8 /** 9 * @param {ListNode} head 10 * @param {number} m 11 * @param {number} n 12 * @return {ListNode} 13 */ 14 var reverseBetween = function(head, m, n) { 15 var res = new ListNode(-1), subHead = res, count = 1; 16 res.next = head; 17 while(count !== m){ 18 subHead = head; 19 head = head.next; 20 count++; 21 } 22 var subTail = head, tmp; 23 while(count !== n + 1){ 24 tmp = head.next; 25 head.next = subHead.next; 26 subHead.next = head; 27 head = tmp; 28 count++; 29 } 30 subTail.next = head; 31 return res.next; 32 };
[LeetCode][JavaScript]Reverse Linked List II
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原文地址:http://www.cnblogs.com/Liok3187/p/4852361.html