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.
分析
如演示样例所看到的。给定一个链表。要求交换链表中相邻两个节点。
对于此题的程序实现，必须注意的是指针的判空，否则。一不注意就会出现空指针异常。
AC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 || head->next == NULL)
return head;
ListNode *p = head , *q = p->next;
//首先交换头两个结点,同一时候保存q后结点
ListNode *r = q->next;
head = q;
p->next = r;
q->next = p;
if (r && r->next)
{
ListNode *pre = p;
p = p->next;
q = p->next;
while (q)
{
//保存q结点后结点
ListNode *r = q->next;
pre->next = q;
p->next = r;
q->next = p;
if (r && r->next)
{
pre = p;
p = r;
q = p->next;
}
else{
break;
}
}
}
return head;
}
};