LeetCode:Swap Nodes in Pairs

题目链接

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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给链表添加一个临时的头结点, 这样操作更方便。其实大部分链表问题,添加一个头结点，都会简化后面的操作   本文地址

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/**  * Definition for singly-linked list.  * struct ListNode {  *     int val;  *     ListNode *next;  *     ListNode(int x) : val(x), next(NULL) {}  * };  */ class Solution { public:     ListNode *swapPairs(ListNode *head) {         ListNode tmphead(0); tmphead.next = head;         ListNode *pre = &tmphead, *p = head;         while(p && p->next)//p 和 p->next是待交换的两个节点，pre是p的前一个节点         {             pre->next = p->next;             p->next = p->next->next;             pre->next->next = p;                           pre = p;             p = p->next;         }         return tmphead.next;     } };

 

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