LeetCode-Swap Nodes in Pairs

题目：

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.

解读：交还给定链表中相邻的两个节点，不得修改节点的值。
注意点：left为左节点，right为右节点，temp为左节点的前一个节点（如果有的话）。left指向right的下一个节点，right指向left，temp（若有）指向right。
temp = left, left = left->next; right = left->next, 
当只剩下一个节点或者没有节点的时候结束。
代码：/**
* 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 == NULL) return head;
if(head->next == NULL) return head;
ListNode* left = head;
ListNode* right = left->next;
ListNode* temp;
ListNode* answer = head;
int flag = 0;
while(right != NULL) {
left->next = right->next;
right->next = left;
if(flag == 0) {
answer = right;
temp = left;
flag = 1;
} else {
temp->next = right;
temp = left;
}
if(left->next == NULL) break;
if((left->next)->next != NULL){
left = left->next;
right = left->next;
} else break;
}
return answer;
}
};