【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.

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* newHead = NULL;
ListNode* frontNode = NULL;
ListNode* midNode = NULL;
ListNode* rearNode = NULL;
if (head == NULL)
return NULL;
//如果只有Head结点，就直接返回Head结点了
if (head->next == NULL)
return head;
frontNode = head->next;
rearNode = head;
//暂存结点，保存rearNode前面的一个node
ListNode* parent = new ListNode(0);
parent->next = rearNode;
//如果链表只有两个结点
if (frontNode->next == NULL)
{
//完成两个结点的交换
rearNode->next = frontNode->next;
frontNode->next = rearNode;
parent->next = frontNode;
return frontNode;
}
//如果链表有不止两个结点
while (rearNode!= NULL && frontNode != NULL)
{
//完成两个结点的交换
rearNode->next = frontNode->next;
frontNode->next = rearNode;
parent->next = frontNode;
//如果head == rearNode，说明这是链表刚开始，要记录下链表头
if (head == rearNode)
head = frontNode;
//如果后面还存在两个（及以上）的结点，就向后移动
if (rearNode->next != NULL && rearNode->next->next != NULL)
{
parent = rearNode;
rearNode = rearNode->next;
frontNode = rearNode->next;
}
else
break;
}
return head;
}
};

收工~