LeetCode刷题笔记（链表）：swap-nodes-in-pairs

Bertha 。 2022-06-01 05:45 109阅读 0赞

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*  **转载请注明作者和出处：[http://blog.csdn.net/u011475210][http_blog.csdn.net_u011475210]**
 *  **代码地址：[https://github.com/WordZzzz/Note/tree/master/LeetCode][https_github.com_WordZzzz_Note_tree_master_LeetCode]**
 *  **刷题平台：[https://www.nowcoder.com/ta/leetcode][https_www.nowcoder.com_ta_leetcode]**
 *  **题  库：Leetcode经典编程题**
 *  **编  者：WordZzzz**

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*   *  题目描述
     *  解题思路
     *  C版代码实现

## 题目描述 ##

Given a linked list, swap every two adjacent nodes and return its head.

For example,  
Given1->2->3->4, you should return the list as2->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.

## 解题思路 ##

通过这么多次的编程练习，链表的顺序交换我相信大家早已轻车熟路。本题中，我们先写一个交换相邻结点的函数swap，此函数交换输入的两个相邻链表结点之后，返回新的第一结点。然后我们开始考虑主函数，同样的新建一个指向头结点的指针（只要头结点有可能被替换，我们既需要这么做），然后遍历整个链表调用交换函数swap即可。

## C++版代码实现 ##

/**
     * Definition for singly-linked list.
     * struct ListNode {
     *     int val;
     *     ListNode *next;
     *     ListNode(int x) : val(x), next(NULL) {}
     * };
     */
    class Solution {
    public:
        ListNode *swap(ListNode *slow, ListNode *fast){
            slow->next = fast->next;
            fast->next = slow;
            return fast;
        }
    
        ListNode *swapPairs(ListNode *head) {
            if(head == NULL || head->next == NULL)
                return head;
            ListNode *dummy = new ListNode(0);
            dummy->next = head;
            ListNode *cur = dummy;
            for(;cur->next != NULL && cur->next->next != NULL; cur = cur->next->next)
                cur->next = swap(cur->next, cur->next->next);
            return dummy->next;
        }
    };

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[http_blog.csdn.net_u011475210]: http://blog.csdn.net/u011475210
[https_github.com_WordZzzz_Note_tree_master_LeetCode]: https://github.com/WordZzzz/Note/tree/master/LeetCode
[https_www.nowcoder.com_ta_leetcode]: https://www.nowcoder.com/ta/leetcode