Leetcode--Reverse Nodes in k-Group

最新推荐文章于 2025-09-06 22:33:46 发布

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最新推荐文章于 2025-09-06 22:33:46 发布

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文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/bhlsheji/p/4093586.html

本文介绍了一种算法，该算法可以将链表中的节点每K个一组进行反转，并返回修改后的链表。通过使用栈来辅助实现节点的反转，确保只使用常量级的额外空间。

Problem Description:

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：看到这个题目第一想法就是就地操作，依次以k个元素为一组来逆置，最后少于k个元素的组不逆置，可是实际操作时将前后两组的首尾相连不是非常好操作，于是直接利用栈将k个元素进栈出栈实现逆置，详细实现时先统计出链表的总元素个数，提前算出须要逆置的组数，代码例如以下：

/**
 * 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 *p=head,*newh=NULL,*pre; 
		stack<ListNode *> stk;
        int len=0,i=1,cnt,n;
        while(p)
        {
            len++;
            p=p->next;
        }
        if(k>len||k<=1)
            return head;
        n=len/k;
        p=head;
        while(i<=n)
        {
            cnt=0;
			while(cnt<k)
			{
				stk.push(p);
				p=p->next;
				cnt++;
			}
			while(!stk.empty())
			{
				if(!newh)
				{
					newh=stk.top();
					pre=newh;
					stk.pop();
				}
				else
				{
					pre->next=stk.top();
					pre=pre->next;
					stk.pop();
				}
			}            
            i++;
        }
        pre->next=p;
        return newh;
    }
};

转载于:https://www.cnblogs.com/bhlsheji/p/4093586.html