Rotate List

一、问题描述

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.

二、思路

常规思路：找到翻转后链表的尾部，然后通过链表的尾部链接到链表的头部即可，返回新链表的头部。

注意考虑边界情况，当

k % size == 0 || size == 1

时不翻转链表。

三、代码

class Solution {
public:
    ListNode* rotateRight(ListNode* head, int k) {
        if( k == 0)
            return head;
        int size = 0;
        ListNode* p = head;
        ListNode* q = head;
        while(p){
            ++size;
            p = p -> next;
        }
        if(size == 0)
            return NULL;
        if(k % size == 0 || size == 1)
            return head;
        p = head;
        int k1 = k % size;
        int end = size - k1;
        while(--end)
            p = p -> next;
        q = p -> next;
        p -> next = NULL;
        p = q;
        while(--k1)
            p = p -> next;
        p -> next = head;
        return q;
    }
};