[LeetCode 007] Reverse Integer

Reverse Integer

• 判断是否overflow的部分注意：
  • 根据x的正负检测。
  • 根据result不准确，我们需要检测x还剩一位数的时候result是否overflow

Implementation

public class Solution {
    public int reverse(int x) {
        int result = 0;
        final int MIN_THRESHOLD = Integer.MIN_VALUE / 10;
        final int MAX_THRESHOLD = Integer.MAX_VALUE / 10;
        while (x != 0) {
            int newResult = result * 10 + x % 10;
            x = x / 10;
            if (x > 0 && (newResult > MAX_THRESHOLD || (newResult == MAX_THRESHOLD && x > 7)))
                return 0;
            if (x < 0 && (newResult < MIN_THRESHOLD || (newResult == MIN_THRESHOLD && x < -8)))
                return 0;
            result = newResult;
        }
        return result;
    }
}

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