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  • LeetCode

     24. Swap Nodes in Pairs

    Problem's Link

     ----------------------------------------------------------------------------

    Mean: 

    给定一个链表,交换这个链表两两相邻的元素.

    analyse:

    Time complexity: O(N)

     

    view code

    /**
    * -----------------------------------------------------------------
    * Copyright (c) 2016 crazyacking.All rights reserved.
    * -----------------------------------------------------------------
    *       Author: crazyacking
    *       Date  : 2016-02-19-10.35
    */
    #include <queue>
    #include <cstdio>
    #include <set>
    #include <string>
    #include <stack>
    #include <cmath>
    #include <climits>
    #include <map>
    #include <cstdlib>
    #include <iostream>
    #include <vector>
    #include <algorithm>
    #include <cstring>
    using namespace std;
    typedef long long(LL);
    typedef unsigned long long(ULL);
    const double eps(1e-8);


    // 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 || head->next==nullptr)
               return head;
           ListNode *frontPtr=head,*backPtr=head->next;
           while(frontPtr && backPtr)
           {
               swap(frontPtr->val,backPtr->val);

               frontPtr=frontPtr->next;
               if(frontPtr!=nullptr)
                   frontPtr=frontPtr->next;

               backPtr=backPtr->next;
               if(backPtr!=nullptr)
                   backPtr=backPtr->next;

           }
           return head;
       }
    };

    int main()
    {

       return 0;
    }
    /*

    */
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  • 原文地址:https://www.cnblogs.com/crazyacking/p/5200178.html
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