# Heading
Tags: algorithms linked-list two-pointers
Langs: c cpp csharp golang java javascript kotlin php python python3 racket ruby rust scala swift typescript
- algorithms
- Medium (40.53%)
- Likes: 395
- Dislikes: -
- Total Accepted: 107.7K
- Total Submissions: 265.3K
- Testcase Example: '[1,2,3,4,5]\n2'
给定一个链表,旋转链表,将链表每个节点向右移动 k 个位置,其中 k 是非负数。
示例 1:
输入: 1->2->3->4->5->NULL, k = 2 输出: 4->5->1->2->3->NULL 解释: 向右旋转 1 步: 5->1->2->3->4->NULL 向右旋转 2 步: 4->5->1->2->3->NULL
示例 2:
输入: 0->1->2->NULL, k = 4 输出:2->0->1->NULL
解释: 向右旋转 1 步: 2->0->1->NULL 向右旋转 2 步: 1->2->0->NULL 向右旋转 3 步:0->1->2->NULL
向右旋转 4 步:2->0->1->NULL
/*
* @lc app=leetcode.cn id=61 lang=javascript
*
* [61] 旋转链表
*/
// @lc code=start
/**
* 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} k
* @return {ListNode}
*/
// var rotateRight = function (head, k) {
// if (!head) return null;
// if (!head.next) return head;
// //连成一个环
// let p = head,length = 1;
// while (p.next) {
// p = p.next
// length++
// }
// let q = p;
// p = head;
// q.next = p;
// //遍历step次,断开环
// const step = length-k%length;
// for (let i = 0; i < step; i++) {
// q = p;
// p = p.next;
// }
// q.next = null;
// return p;
// };
var rotateRight = function (head, k) {
if (!head) return null;
let p = head, length = 0, q;
//计算length,q保存尾结点
while (p) {
q = p;
p = p.next;
length++;
}
if(k%length === 0) return head;
//找到头结点的前一个节点
let preHeadIdx = length - k % length - 1;
p = head;
while (preHeadIdx-- > 0) {
p = p.next;
}
//断开并用尾结点连接head
const newHead = p.next;
p.next = null;
q.next = head;
return newHead;
};
// @lc code=end
// @after-stub-for-debug-begin
module.exports = rotateRight;
// @after-stub-for-debug-end
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