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24. Swap Nodes in Pairs

时间:2016-09-12 14:17:43      阅读:137      评论:0      收藏:0      [点我收藏+]

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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         
13         if(head == NULL || head->next == NULL){
14             return head;
15         }
16         ListNode* newHead = new ListNode(-1);
17         ListNode* tail = newHead;
18         newHead->next = head;
19         
20         while(tail && tail->next && tail->next->next){
21             ListNode*  n = tail->next;
22             ListNode*  nn = tail->next->next;
23             
24             //tail->next = nn;
25             n->next = nn->next;
26             nn->next = n;
27             tail->next = nn;
28             
29             tail = tail->next->next;
30         }
31         
32         return newHead->next;
33     }
34 };

 

24. Swap Nodes in Pairs

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原文地址:http://www.cnblogs.com/sankexin/p/5864426.html

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