61. Rotate List

Given a list, rotate the list to the right by k places, where k is non-negative.

Example:

Given 1->2->3->4->5->NULL and k = 2,return 4->5->1->2->3->NULL.

思路：首先遍历链表算出链表长度len，然后计算出k%len，即所需要的最小rotate次数，然后把链表分成两段交换即可。

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* rotateRight(ListNode* head, int k) {        if (!head) return NULL;        //计算链表的长度        ListNode* tmp = head;        int len = 0;        while (tmp) {            len++;            tmp = tmp->next;        }        //旋转k%len次得到的结果和旋转k次得到的结果是相同的        k = k % len;        if (k == 0) return head;        //post->next指向的是倒数第k个元素        ListNode* pre  = head;        ListNode* post = head;        for (int i = 0; i < k; i++) {            pre = pre->next;        }        while (pre->next) {            pre = pre->next;            post = post->next;        }        //post->next前后的链表互换位置，即得到最终结果        ListNode* newHead = post->next;        pre->next = head;        post->next = NULL;        return newHead;    }};