reverse-integer

题目描述
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?

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).
重点：判断溢出

public class Solution {
    public int reverse(int x) {
        int res=0;
        while (x!=0){
            int tail=x%10;
            int newRes=res*10+tail;
            if((newRes-tail)/10!=res)如果newRes-tail)/10!=res说明产生了溢出
                //当res*10+tail的值超过最大位数时，系统会将超过最大位数的高位二进制数扔掉，
                //从而导致了newRes!=res*10+tail.此时计算的(newRes-tail)/10自然也不等于res了
                return 0;
            res=newRes;
            x=x/10;
        }
        return res;
    }
}

class Solution:
    def reverse(self, x: int) -> int:
        flag = -1 if x < 0  else 1
        res = flag * int(str(abs(x))[::-1])
        return res if (-2**31)<=res<=(2**31-1) else 0