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：

/**
* 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 dummy(0);
dummy.next = head;
head = &dummy;
while (head -> next != nullptr && head -> next -> next != nullptr) {
ListNode* temp1 = head -> next;
head -> next = head -> next -> next;
ListNode* temp2 = head -> next -> next;
head -> next -> next = temp1;
temp1 -> next = temp2;
head = head -> next -> next;
}
return dummy.next;
}
};