7.Reverse Integer

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Question：

Reverse digits of an integer.

Example1: x = 123, return 321

Example2: x = -123, return -321

Example3: x = 120, return 21

Note：Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

My answer in Java：

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

Better answer in Java：

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class Solution {
public:
    int reverse(int x) {
        long long res = 0;
        while (x != 0) {
            res = 10 * res + x % 10;
            x /= 10;
        }
        return (res > INT_MAX || res < INT_MIN) ? 0 : res;
    }
};

Didn’t thought about over integer overflows, why this problem will happen, we know that the range of integer is -2147483648～2147483647. So if we want to traverse the integer 1000000009, the traversed number 9000000001 will be out of range, so we need to judge the overflow situation.

Answer in python:

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class Solution(object):
    # def reverse(self, x):
    #     """
    #     :type x: int
    #     :rtype: int
    #     """
    #     res=0
    #     if x==0:
    #         return 0
    #     while x!=0:
    #         res=res*10+x%10
    #         x=x/10
    #     return res if -2147483648 <= res <= 2147483647 else 0
        
        
   def reverse(self, x):
       s = str(x)
       res = int('-' + s[1:][::-1]) if s[0] == '-' else int(s[::-1])
       return res if -2147483648 <= res <= 2147483647 else 0

But I have a consider is that why the code which are annotated can’t not run when the input integer is -321