leetcode 25. Reverse Nodes in k-Group

最新推荐文章于 2022-04-08 22:28:12 发布

原创最新推荐文章于 2022-04-08 22:28:12 发布 · 270 阅读

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本文介绍了一种链表操作的方法——K组翻转。该方法通过对链表节点进行特定的翻转操作来实现每K个一组的逆序排列。文章详细解释了如何通过迭代方式完成这一过程，并提供了一个具体的例子帮助理解。

Discription

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked 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

思路

模拟题

时间复杂度： $?$
空间复杂度： $?$

代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reverse(ListNode* p1, ListNode* p4)
    {
        ListNode *pre = NULL, *cur = p1, *next;
        while (pre != p4)
        {
            next = cur->next;

            cur->next = pre;
            pre = cur;
            cur = next;
        }
    }
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode* pVirtualHead = new ListNode(-1);
        pVirtualHead->next = head;

        ListNode *p0 = pVirtualHead, *p1, *p3, *p4;
        while (true)
        {
            p1 = p0->next;
            p3 = p0;

            for (int i = 0; i < k; i++)
            {
                if (!p3->next) return  pVirtualHead->next;
                p3 = p3->next;
            }
            p4 = p3->next;

            reverse(p1, p3);
            p0->next = p3;
            p1->next = p4;

            p0 = p1;
        }
    }
};