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.
代码不是最简的,处理head的部分明显可以和后面合并。关键记住无论思路多清晰都需要模拟一下。
/**
* 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 *first = head;
if(head == NULL) return NULL;
ListNode *second = head -> next;
ListNode *temp;
if(second != NULL)
{
head = second;
temp = first;
first -> next = second -> next;
second -> next = first;
first = first->next;
if(first == NULL ||first -> next == NULL )return head;
second = first ->next;
temp ->next = second;
}
while(first != NULL && second != NULL)
{
temp = first;
first -> next = second -> next;
second -> next = first;
first = first->next;
if(first == NULL ||first -> next == NULL )return head;
second = first ->next;
temp ->next = second;
}
return head;
}
};