24. Swap Nodes in Pairs
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.
题目大意：
交换每两个节点的位置。
代码如下：
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/**
* 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* left,*right,*pre,*p;
pre = NULL;
//记录每两个节点前面的那个节点
p = head;
while
(p !=NULL && p->next != NULL)
{
left = p;
right = p->next;
left->next = right->next;
right->next = left;
if
(pre != NULL)
{
pre->next = right;
}
else
//链表的头两个节点交换位置
{
head = right;
}
pre = left;
p = left->next;
}
return
head;
}
};
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本文转自313119992 51CTO博客，原文链接:http://blog.51cto.com/qiaopeng688/1837471
