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

假设有连续的4个节点A->B->C->D，其中A为已经交换过的上一轮的节点。则一轮交换后为A->C->B->D。运用链表的操作方法实现这些节点关系的改变即可。

实现

/**
 * 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 dummy(-1);
        ListNode *p=&dummy, *tmp;
        dummy.next = head;
        while(p!=NULL && p->next!=NULL && p->next->next!=NULL){
            tmp = p->next->next;
            p->next->next = tmp->next;
            tmp->next = p->next;
            p->next = tmp;
            p = p->next->next;
        }
        return dummy.next;
    }
};

思考

对于这题来说，使用三个指针代码会简洁一些。