LeetCode 7. Reverse Integer (Java)

tags: leetcode Java

Description

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.
Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

將整數反轉，遇到超過32-bit的數就回傳0

Example

Input: x = 123
Output: 321

Input: x = -123
Output: -321

Input: x = 0
Output: 0

Idea

• 要用到Integer.MAX_VALUE及Integer.MIN_VALUE去防止32-bit以上的數回傳錯誤

Code

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

Result

Runtime: 1 ms, faster than 100.00% of Java online submissions for Reverse Integer.
Memory Usage: 35.6 MB, less than 98.93% of Java online submissions for Reverse Integer.