【LeetCode】Reverse Nodes in k-Group

题目描述：

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

这题算是前一题的升级版。难点在两个：一是用常数空间解，二是最后剩余长度不足k时顺序不变。

如果无视常数空间这个要求的话，可以用个vector<ListNode*>(k)写个循环来很方便地解。

如果无视第二个要求的话，直接将一个节点插入到前一组的最后一个节点后面就好了，因为多了这个条件所以我们不得不将整个链表遍历两遍来确定长度。不知道有没有只遍历一遍的方法。

代码：

class Solution {
public:
	ListNode *reverseKGroup(ListNode *head, int k) {
		ListNode *res = new ListNode(0);
		res->next = head;
		ListNode *prev = res;
		int count(0);
		while (head){
			count++;
			head = head->next;
			if (count >= k){
				prev = revers(prev, prev->next, k);
				prev->next = head;
				count = 0;
			}
		}
		return res->next;
	}
	ListNode* revers(ListNode *prev, ListNode *head, int k){
		int count(0);
		ListNode *res = head;
		while (head&&count < k){
			ListNode *next = prev->next;
			ListNode *curr = head;
			head = head->next;
			prev->next = curr;
			curr->next = next;
			count++;
		}
		return res;
	}
};