LeetCode Rotate List

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.

题意：翻转链表。

思路：注意这里的k可能是大于n的，所以要先进行取模处理，然后就是：把链表头尾相连，然后就可以翻转了。

/**
 * 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) return head;

		ListNode *tail = head;
		int n = 1;
		while (tail->next != NULL) {
			n++;
			tail = tail->next;
		}
		k %= n;
		if (k == 0) return head;

		int len = n - k;
		ListNode *cur = head;
		while (--len) {
			cur = cur->next;
		}
		ListNode *start = cur->next;
		cur->next = NULL;
		tail->next = head;
		return start;
    }
};

LeetCode Rotate List

原文：http://blog.csdn.net/u011345136/article/details/44620437