Rotate List

题目大意

给出一个单链表，和一个K值，根据K值往右旋转，例如：

思路

先求出链表的长度size，其实按着倒数第K%size个位置旋转，这个位置即size-(K%size)

参考代码

/**

 * 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 == NULL || k < 0)

            return NULL;

        ListNode *tmp = head;

        int size = 0;

        while(tmp != NULL)

            {

                tmp = tmp->next;

                ++size;

            }

        k = size - (k % size);

        if(k == size)

            return head;

        k = k - 1;

        ListNode *rev = NULL;

        ListNode *pre = head;

        ListNode *token = head;

        ListNode *cur = head;

        int index = 0;

        while(index <= k && cur != NULL)

        {

            pre = cur;

            cur = cur->next;

            ++index;

        }

        rev = cur;

        if(cur == NULL)

            return head;

        while(cur->next != NULL)

            cur = cur->next;

        cur->next = token;

        pre->next = NULL;

        return rev;

    }

};

注意

两段链表接在一起的时候，注意原来前半部分最后一个指针应当赋值为NULL