LeetCode - 7. Reverse Integer 7. Reverse Integer Problem's Link

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Mean: 

将一个整数的数值位反转.

analyse:

题目没说当精度溢出时返回0.这个地方要注意一下.

Time complexity: O(N)

view code

/**

* -----------------------------------------------------------------

* Copyright (c) 2016 crazyacking.All rights reserved.

*       Author: crazyacking

*       Date  : 2016-02-15-15.08

*/

#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);

class Solution

{

public:

   int reverse(int x)

   {

       bool isNeg(x<0?1:0);

       int64_t xx;

       if(x<0) xx=-1*(int64_t)x;

       else xx=(int64_t)x;

       int64_t ret=0;

       while(xx)

       {

           int tmp=xx%10;

           xx/=10;

           ret=ret*10+tmp;

       }

       if(ret>INT_MAX)

           return 0;

       if(isNeg)

           ret=-ret;

       return ret;

   }

};

int main()

   cout<<LLONG_MAX<<endl;

   Solution solution;

   int x;

   while(cin>>x)

       cout<<solution.reverse(x)<<endl;

   return 0;

}