Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
Input: 120
Output: 21
Note:
Assume we are dealing with an environment which could only store integers within the 32-bit signed integer range: [−2^31, 2^31 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
思路：按照除留余数的方法进行转换，但是要考虑到例外情况，如果转换之后的数超出了INT的范围，就要返回0，所以在最开始定义result变量的时候，把它定义为long，确保转换之后不会出现错误，最后再返回结果的时候判断一下是否大于INT的范围，是则返回0，否则强制转换为int再返回
package com.Ryan;
public class ReverseInteger {
public static void main(String[] args) {
ReverseInteger reverseInteger = new ReverseInteger();
int a = -123;
System.out.println(reverseInteger.reverse(a));
}
public int reverse(int x) {
long result = 0;
while (x != 0) {
result = result * 10 + x % 10;
x = x / 10;
}
if (result > Integer.MAX_VALUE || result < Integer.MIN_VALUE) {
return 0;
} else
return (int) result;
}
}