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;
}