leetcode025：Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

问题分析

• ans是最终结果的表头；
• ans_p负责组间的连接工作；
• p负责检查是否有k个结点逆转；
• p1用来记录逆转前的组的表头；
• q为逆转过程中的表头。

//运行时间：<span style="font-family:Helvetica Neue, Helvetica, Arial, sans-serif;color:#333333;"><span style="font-size: 14px; line-height: 20px; background-color: rgb(249, 249, 249);">38ms</span></span>

class Solution {
public:
	ListNode *reverseKGroup(ListNode *head, int k) {
		if (!head) return NULL;
		if (k<=1) return head;
		ListNode *ans = NULL;
		ListNode *ans_p = head;
		ListNode *p = head;//前进！
		ListNode *p1;
		ListNode *q = head;
		ListNode *x;
		ListNode *y;
		int count ;
		while (1){
			count = 0;
			p1 = p;
			while (p&&count < k){
				count++;
				p = p->next;
			}
			if (count < k) {
				if (!ans)	ans = p1; 
				else{ ans_p->next = p1; }
				break;
			}
			else{
				x = y = q->next;
				count = 1;
				while (count < k){
					x = x->next;
					y->next = q;
					q = y;
					y = x;
					count++;
				}
				if (!ans) { ans = q; ans_p = p1; }
				else { ans_p->next = q; ans_p = p1; }
				q = p;
			}
			if (!p) break;
		}
		return ans;
	}
};