LeetCode - 24. Swap Nodes in Pairs 24. Swap Nodes in Pairs Problem's Link

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Mean: 

给定一个链表，交换这个链表两两相邻的元素.

analyse:

略

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;

}

/*