LeetCode - 24. Swap Nodes in Pairs

 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:

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;
}
/*

*/

转载于:https://www.cnblogs.com/crazyacking/p/5200178.html