Q：

Reverse digits of an integer.
Example1: x = 123, return 321Example2: x = -123, return -321

• 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?
• For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

A:

public class Solution {
    public int reverse(int x) {
        long result =0;     //为了处理Integer的边界值，这里初始化成long
        while(x != 0)
        {   //之前的结果乘以10（在前，前一位），后来的值是module 10求模得到的（在后，后一位）
            result = (result*10) + (x%10); 
            if(result > Integer.MAX_VALUE) return 0;
            if(result < Integer.MIN_VALUE) return 0;
            x = x/10;       //处理下一位，从右至左
        }
        return (int)result; 
    }
}

test case: 939

• result = 0*10 + 939%10 = 0 + 9 = 9
  x = 939/10 = 93
• result = 9*10 + 93%10 = 90 + 3 = 93
  x = 93/10 = 9
• result = 93*10 + 9%10 = 939
  x = 9/10 = 0 (结束)

test case: -100

• result = 0*10 + (-100)%10 = 0 + 0 = 0
  x = (-100)/10 = (-10)
• result = 0*10 + (-10)%10 = 0 + 0 = 0
  x = (-10)/10 = -1
• result = 0*10 + (-1)%10 = 0 + (-1) = -1
  x = -1/10 = 0 (结束)

Notes:

关于module取模

(-10)/3 = -3
10/(-3) = -3
(-10)/(-3) = 3

(-10)%3 = -1
10%(-3) = 1
(-10)%(-3) = -1

5%3 = 2
-5%3 = -2
5%-3 = 2
-5%-3 = -2 //结果符号由之前的那个数字来决定
5.3%3 = 2.3
5%3.3 = 1.7000000000000002 //取模支持小数
5.3%3.3 = 2.0 //两个小数也可以取模