【leetcode】Swap Nodes in Pairs

最新推荐文章于 2024-10-10 01:02:39 发布

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最新推荐文章于 2024-10-10 01:02:39 发布

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原文链接：http://www.cnblogs.com/jawiezhu/p/4452970.html

本文介绍了一种在常数空间复杂度下交换链表中每两个相邻节点的算法。通过递归方式实现，不修改链表值仅变更节点连接，示例展示了1->2->3->4经算法处理后变为2->1->4->3的过程。

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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.

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode *swapPairs(ListNode *head) {
12        if (head == NULL)
13             return NULL;
14        if (head -> next == NULL)
15             return head;
16        ListNode *tmp = head -> next;
17        head -> next = swapPairs(tmp -> next);
18        tmp -> next = head;
19        return tmp;
20     }
21 };