61. Rotate List Add to List Share Given a linked list, rotate the list to the right by k places, where k is non-negative. Example 1: Input: 1->2->3->4->5->NULL, k = 2 Output: 4->5->1->2->3->NULL Explanation: rotate 1 steps to the right: 5->1->2->3->4->NULL rotate 2 steps to the right: 4->5->1->2->3->NULL Example 2: Input: 0->1->2->NULL, k = 4 Output: 2->0->1->NULL Explanation: rotate 1 steps to the right: 2->0->1->NULL rotate 2 steps to the right: 1->2->0->NULL rotate 3 steps to the right: 0->1->2->NULL rotate 4 steps to the right: 2->0->1->NULL
leetcode 61 题，给到一个链表， 顺时针旋转k个，求结果
解题思路，移动k个，新建两个指针，让前一个指针前进k个格子，然后后一个指针再同时前进，这样就可以找到切割点了。
需要注意的是，k有可能是会超出链表长度的， 需要进行处理，还有就是注意一些特殊情况的排除
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ /* 本题是说将一串链表顺时针旋转k个， */ func rotateRight(head *ListNode, k int) *ListNode { if k == 0 || head==nil { return head } var ptr1, ptr2 *ListNode = head, head // var ptr2 *ListNode = head var i int = 0 // first try to find the diff node length for ; i < k; i ++ { if ptr1.Next == nil { break } else { ptr1 = ptr1.Next } } // the k is to big, i+1 is the length of the list node if i < k { i = k % (i + 1) return rotateRight(head, i) } // to move to the tail for { if ptr1.Next == nil { break } else { ptr1 = ptr1.Next ptr2 = ptr2.Next } } //do split ptr1.Next = head head = ptr2.Next ptr2.Next = nil return head }
原文：https://www.cnblogs.com/mangmangbiluo/p/13196900.html