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
题目:把给定链表中的相邻的两个节点交换翻转,并返回;
例如:1->2->3->4 输出: 2->1->4->3;
注意:涉及到指针的操作,注意前两个节点在交换的时候要注意head指针的变化。
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 NULL;
ListNode *p = head;
ListNode *q = p->next;
if(q==NULL) return head;
int flag = 0;
while(q!=NULL&&p!=NULL)
{
if(flag == 0)//前两个节点交换
head = q;
p->next = q->next;
q->next = p;
ListNode *p1 = p;
p=p->next;
if(p==NULL)
break;
else
{
q=p->next;
if(q!=NULL)
p1->next = q;
flag = 1;
}
}
return head;
}
};