[LeetCode] Reverse Integer

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?

Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).

以下代码没有考虑溢出，只考虑了负值和反转后开始为0的情况：

class Solution {
public:
    int reverse(int x) {
        int y =x;
        int res=0;
        while(y){     
           y   = y/10;
           res  =  res*10+y%10;
        }//end while
        return res;
    }//end reverse
};

考虑溢出的情况，由于在32位机和64位机上，sizeof(int)都是4；所以int值即带符号的int值的最大值是7fffffff，

考虑x有正有负的情况。先把x变成正的，进行判断溢出与否，没有溢出则输出正常的反转后int值。

代码如下：

class Solution {
public:
    int reverse(int x) {
        bool neg=false;
        if (x < 0) neg = true, x = -x;
        unsigned int ret = 0;
        while (x ){
            ret = ret*10+ x%10;
            x = x / 10;
        }

        if (ret>0x7fffffff) return 0; //overflow check.
        if (neg) return -(int)ret;
        return (int)ret;
    }
};