[LeetCode] #25 Reverse Nodes in k-Group

最新推荐文章于 2025-09-12 17:31:31 发布

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原文链接：http://www.cnblogs.com/Scorpio989/p/4549794.html

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#数据结构与算法

本文介绍了一种链表反转算法，针对链表中的节点以K为单位进行反转操作，同时保留了非K节点的原始顺序。算法通过定义辅助函数判断链表长度是否满足K个节点，并在满足条件时执行反转操作。实现中使用常量内存，确保高效处理链表结构。

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

每次取k个长度的链表进行反转。时间：30ms

代码如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool isKLen(ListNode* head, int k){
        if (k <= 0 || head == NULL)
            return false;
        ListNode *phead = head;
        int count = k - 1;
        while (phead->next != NULL&&count){
            phead = phead->next;
            count--;
        }
        if (count == 0)
            return true;
        else
            return false;
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        if (head == NULL || k <= 1)
            return head;
        ListNode *p, *q, *phead, *pnext = new ListNode(0);
        pnext->next = p = head;
        phead = pnext;
        while (isKLen(p,k)){
            int count = k-1;
            ListNode *qt = pnext->next;
            while (count--){
                q = p->next;
                p->next = q->next;
                q->next = qt;
                qt = q;
            }
            pnext->next = qt;
            pnext = p;
            p = p->next;
        }
        return phead->next;
    }
};