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Given a list, rotate the list to the right by k places, where k is non-negative.
For example:
Given 1->2->3->4->5->NULL and k = 2,
return 4->5->1->2->3->NULL.
分析：首先遍历list得到长度，然后找到应该rotate的位置，进行rotate。注意：k有可能大于长度，因此要进行取模运算。
运行时间：12ms
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* rotateRight(ListNode* head, int k) { 12 if(!head || !head->next) return head; 13 if(k == 0) return head; 14 15 int len = 0; 16 ListNode* move = head; 17 while(move->next){ // find the end of the list 18 len++; 19 move = move->next; 20 } 21 ListNode* end = move; 22 len++; // length of the list 23 24 int count = 0; 25 move->next = head; 26 while(count < len - k % len){ // find the position of the rotating node 27 count++; 28 move = move->next; // move is the position of the node before the rotating node 29 } 30 ListNode* rotate = move->next; // rotate is the position of the rotating node 31 32 end->next = head; 33 move->next = NULL; 34 return rotate; 35 } 36 };
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原文地址：http://www.cnblogs.com/amazingzoe/p/4512809.html
