Reverse Integer [LeetCode]

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).

 

Solution: be careful about corner case, like INT_MIN, 0, and -INT_MIN is not equals to INT_MAX.  Here is the definition of them in limits.h

#define INT_MIN     (-2147483647 - 1) /* minimum (signed) int value */
#define INT_MAX       2147483647    /* maximum (signed) int value */

 

    int reversePostiveInt(int x){
        vector<int> digits;
        while(true) {
            if(x < 10 ){
                digits.push_back(x);
                break;
            }else {
                int remainder = x % 10;
                int quotient = x / 10;
                digits.push_back(remainder);
                x = quotient;
            }
        }
        // reverse output
        long long int ret = 0;
        for(int i = 0; i < digits.size(); i ++) {
            ret = ret * 10 + digits[i];
            if(ret > INT_MAX){
                ret = INT_MAX;
                break;
            }
        }
        return (int) ret;
    }
    
    int reverse(int x) {
        if(x == 0 )
            return 0;
        if(x > 0) {
           return reversePostiveInt(x);
        }else if( x == INT_MIN){
            //overflow
            return INT_MIN;
        }else if ( x > INT_MIN) {
            return -reversePostiveInt(-x);
        }
    }