[Leetcode]Swap Nodes in Pairs

最新推荐文章于 2024-10-10 01:02:39 发布

泉山绿树

最新推荐文章于 2024-10-10 01:02:39 发布

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本文介绍了一种链表操作算法——成对交换相邻节点的方法，并提供了完整的C++实现代码。该算法仅使用常数空间，不修改链表值而只变更节点连接，通过迭代方式完成节点交换。

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Swap Nodes in Pairs My Submissions Question
Total Accepted: 69767 Total Submissions: 210542 Difficulty: Medium
Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* swapPairs(ListNode* head) {
        ListNode* h = new ListNode(-1);
        h->next = head;
        auto p = h;
        ListNode* begin;
        ListNode* next;
        while(p && p->next && p->next->next){ //should think the next and next's next
            begin = p->next;
            next = begin->next;
            p->next = next;
            begin->next = next->next;
            next->next = begin;
            p = begin;
        }
        return h->next;
    }
};

又是一遍AC。。。链表的细节实现题只要在纸上画好了基本没问题

题数也达到了90~