【Leetcode】之Swap Nodes in Pairs

一.问题描述

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.

二.我的解题思路

这种链表的题一向不难解决，在纸上写写画画往往就知道该怎么求解。我是每两个节点进行一次操作，设立了四个指针。这四个指针分别指向，当前节点上一个节点（pri），当前节点(curr)，下一个节点(next)，下下个节点(tmp)。测试通过的程序如下：

/**
 * 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) {
        ListNode * curr,* next,* tmp,*res,*pri;
        if(head==NULL)
            return NULL;
        if(head->next==NULL)
            return head;
        curr=head;next=curr->next;
            /*first done*/
          tmp=next->next;
          next->next=curr;
          curr->next=tmp;
          res=next;
          pri=curr;
          curr=curr->next;

        while(curr!=NULL){
            next=curr->next;
            if(next==NULL)
                break;
            if(next->next!=NULL){
                tmp=next->next;
                next->next=curr;
                curr->next=tmp;
                curr=curr->next;
                pri->next=next;
                pri=next->next;
            }
            else{
                pri->next=next;
                next->next=curr;
                curr->next=NULL;
            }


        }
        return res;


    }
};