[leetcode 7] Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Note:
Assume we are dealing with an environment that could only store integers within the 32-bit signed integer range: [−231,  231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

 

Example 1:

Input: x = 123 Output: 321

Example 2:

Input: x = -123 Output: -321

Example 3:

Input: x = 120 Output: 21

Example 4:

Input: x = 0 Output: 0

 

Constraints:

• -231 <= x <= 231 - 1

문제 풀이:

easy인데 많이 틀렸다 ㅎㅎ

reversed된 숫자가 범위를 넘는지 체크하는게 어려웠다.

아래 처럼 result를 int로 선언해서 체크하는 방법도 있고 

long으로 선언해서 마지막에 체크하는 방법도 있다.

class Solution {
    public int reverse(int x) {
        
        int result = 0;
        
        while(x != 0){
            int tmp = x % 10;
            x /= 10;

            if(result > Integer.MAX_VALUE/10 || 
               (result == Integer.MAX_VALUE / 10 && tmp > 7 )) return 0;
            
            if(result < Integer.MIN_VALUE/10 || 
               (result == Integer.MIN_VALUE / 10 && tmp < -8 )) return 0;
            
            result = result*10 + tmp;
        }
        
        return result;
    }
}

 

long 선언

class Solution {
    public int reverse(int x) {
        
        long result = 0;
        
        while(x != 0){
            int tmp = x % 10;
            x /= 10;
            
            result = result*10 + tmp;
            
        }
        
        if(result > Integer.MAX_VALUE || result<Integer.MIN_VALUE){
            return 0;
        }
        
        return (int)result;
    }
}