LeetCode

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.

题意：交换一个链表中相邻的两个结点。
先存一下next和next->next结点，这里有个交换顺序在
0 1 2 3 4（0为我们附加在头结点前的结点）
先将1指向3，再讲0指向2，最后将2指向1。

/** * 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) {        if (!head || !head->next) return head;        ListNode* ans = new ListNode(0);        ans->next = head;        ListNode* cur = ans;        while (cur->next && cur->next->next) {            ListNode* first = cur->next;            ListNode* second = cur->next->next;            first->next = second->next;            cur->next = second;            cur->next->next = first;            cur = cur->next->next;        }        return ans->next;    }};