LeetCode24 Swap Nodes in Pairs

题意:

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.(Easy)

 

分析:

链表题,画个图就比较清晰了,就是每两个进行一下翻转,注意翻转部分头尾指针的指向即可,头指针会变,所以用下dummy node。

代码:

 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         ListNode dummy(0);
13         dummy.next = head;
14         head = &dummy;
15         while (head -> next != nullptr && head -> next -> next != nullptr) {
16             ListNode* temp1 = head -> next;
17             head -> next = head -> next -> next;
18             ListNode* temp2 = head -> next -> next;
19             head -> next -> next = temp1;
20             temp1 -> next = temp2;
21             head = head -> next -> next;
22         }
23         return dummy.next;
24     }
25 };

 

 

 
posted @ 2016-08-15 19:11  wangxiaobao1114  阅读(133)  评论(0编辑  收藏  举报