/**

* Created by lichenxi on 2017/11/12.

* 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 hold integers within the 32-bit signed integer range.

For the purpose of this problem,

assume that your function returns 0 when the reversed integer overflows.

输入123 输出321 反转一下 -123输入 输入-321 如果是120 反转后就是21 0要忽略

123求余 得到3 3*10+2=32 32*10+1=321 ok得到

120

题目提到了溢出问题 只能把数保持在一个integer的整数范围内 也就是我们要保证反转之后 数字要保持在integer范围内

我为什么要把int转成long类型 这个问题 不要问我

*/

public class ReverseInteger {

public int reverse(int x) {

long temp=x;

long result=0;

while (temp!=0){

result=result*10+temp%10;

temp=temp/10;

}

if (result>Integer.MAX_VALUE|| result<Integer.MIN_VALUE){

result=0;

}

return (int)result;

}

}