leetcode 25. Reverse Nodes in k-Group-优快云博客

本文链接：https://blog.youkuaiyun.com/csdn_lisword/article/details/50978737

本文介绍了一种算法，该算法将链表中的节点每K个一组进行反转，并返回修改后的链表。讨论了当节点数量不是K的倍数时如何处理剩余节点，以及如何在不改变节点值的情况下仅改变节点本身来实现这一操作。

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1.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

将一个链表一次反转k个节点直到结束。

2.思路

用一个计数器count,每当计数器count的值达到k,就将这k个节点反转；

要反转这k个节点，需要知道这k个节点的头节点和尾节点，分别用start和end表示，还需要知道这k个节点的前一个节点，以完成前后链表的链接，用pre表示，代码如下。

class Solution {
public:
    ListNode* reverseK(ListNode* start, ListNode* end,ListNode* pre) { // 反转以start开头，end结束的k个节点，返回反转后k个节点的尾节点
		ListNode* tail = start;
		ListNode *tmp;
		while (start != end){
			tmp = start->next;
			start->next = end->next;
			end->next = start;
			start = tmp;
		}
		pre->next = end;
		return tail;
	}
	ListNode* reverseKGroup(ListNode* head, int k) 	{
		ListNode *newhead = new ListNode(-1);
		ListNode *pre = newhead,*cur = head,*start = head,*end = head;
		int count = 1;

		while (cur){
			if (count == 1)	start = cur;
			if (count == k){
				end = cur;
				cur = start;
				pre = reverseK(start, end, pre);
				count = 0;
			}
			cur = cur->next; count++;
		}
		return newhead->next;
	}
	};