LeetCode 25.Reverse Nodes in k-Group

题目:

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

分析与解答:是上个题目的扩展,这次需要一次性反转K个相邻节点,那么可以考虑用栈,将K个节点顺序进栈然后再出栈,就完成了一次反转。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *reverseKGroup(ListNode *head, int k) {
  	ListNode *result, *p, *q;
	result = q = new ListNode(0);
	p = head;
	q->next = head;
	vector<ListNode *> s;
	int count = k;
	while (p != NULL) {
		s.push_back(p);
		p = p->next;
		count--;
		if (count == 0) { //进行反转
			while (count < k) {
				ListNode *temp = s.back();
				s.pop_back();
				q->next = temp;
				q = q->next;
				count++;
			}
			q->next = p;
		}
	}
	return result->next;
    }
};


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