【leetcode每日一题】24.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.

解析：定义三个节点temp、first和second，temp指向上一节点，first指向当前节点，second指向下一节点。让temp指向second，first指向second的下一节点，second指向first。然后依次遍历整个链表即可。代码如下：

/**
 * 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) {
        if(head==NULL||head->next==NULL)
            return head;
        ListNode *first=head;
        ListNode *second=first->next;
        ListNode *temp=first;
        head=second;
        while(first!=NULL&&second!=NULL)
        {
            temp->next=second;
            first->next=second->next;
            second->next=first;
            temp=first;
            first=first->next;
            if(first!=NULL)
                second=first->next;
        }
        return head;
    }
};