LeetCode OJ 24. Swap Nodes in Pairs-优快云博客

本文介绍了一种链表中相邻节点的交换算法实现，通过迭代方式完成每两个节点的交换，同时考虑了奇数和偶数节点的情况。该算法仅使用常数空间，并且不修改链表中的值，只改变节点之间的连接。

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.

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解答：

这道题目的话就是区分奇数个节点和偶数个节点，然后注意一下只有一个头节点、空链表的情况就好了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* swapPairs(struct ListNode* head) {
    struct ListNode *tmp, *new_head = head, *tail = NULL;
    
    while(NULL != head&&NULL != head->next){
        tmp = head->next;
        head->next = head->next->next;
        tmp->next = head;
        if(head == new_head){
            new_head = tmp;
        }
        else{
            tail->next = tmp;
        }
        tail = head;
        head = head->next;
    }
    if(NULL != head&&NULL != tail){
        tail->next = head;
    }
    return new_head;
}