Leetcode | Swap Nodes in Pairs

最新推荐文章于 2024-10-10 01:02:39 发布

转载最新推荐文章于 2024-10-10 01:02:39 发布 · 46 阅读

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原文链接：http://www.cnblogs.com/linyx/p/4036498.html

本文介绍了一种链表中相邻节点的交换算法实现，通过仅使用常数空间完成节点的交换，而不会修改链表中的值。该算法适用于1->2->3->4这样的链表输入，能将其转换为2->1->4->3的形式。

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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.

比起第一次写简洁多了。

 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) return head;
13         ListNode h(0), *first = head, *second, *prev = &h, *nextFirst;
14         h.next = head;
15         
16         while (first) {
17             second = first->next;
18             if (!second) break;
19             prev->next = second;
20             nextFirst = second->next;
21             second->next = first;
22             first->next = nextFirst;
23             prev = first;
24             first = nextFirst;
25         }
26         
27         return h.next;
28     }
29 };