Reverse Integer

原题目   https://leetcode.com/problems/reverse-integer/

Reverse digits of an integer.

Example1: x = 123, return 321
Example2: x = -123, return -321

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Have you thought about this?

Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!

If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.

Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?

For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

Update (2014-11-10):
Test cases had been added to test the overflow behavior.

题目大意，反转int数字

#include <iostream>
using namespace std;


//反转int （利用了long long）
int reverse(int x) {
    long long s = 0;         
    while (x != 0) {
        s = s * 10 + x %10;
        x = x / 10;
    }
    if (s > INT_MAX || s < INT_MIN)
        return 0;
    return s;
}

//反正int （没有使用long long）
int reverse(int x){
    int s = 0;
    while(x != 0) {
        // s * 10 大于Int_Max 或者 s * 10 + x % 10 大于Int_Max
        if (x > 0 && (s > INT_MAX / 10 || (s == INT_MAX / 10 && x % 10 > INT_MAX % 10))) 
            return 0;
        if (x < 0 && (s < INT_MIN / 10 || (s == INT_MIN / 10 && x % 10 < INT_MIN % 10))) 
            return 0;
        s = s * 10 + x % 10;
        x /= 10;
    }
    return s;
}

void test(){
    int x = 4568412;
    // x = -4568412;
    // x = 1000;
    // x = -1000;
    // x = 0;
    // x = INT_MAX + 1;
    // x = INT_MIN - 1;
    reverse(x);
}

int main()
{
    
    test();

    system("pause");
    return 0;
}