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: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
由题意很简单的推出：将数字反转，但提示中有说道 反转后的数字 如果是超过了 int的最大值或是最小值 则返回0，这里我们可以利用long进行返回 然后long转int，long的位数比int大，只需判断long的结果是否大于或小于int的最大最小值即可最终代码如下：
public int reverse(int x) {
long result = 0;
int tmp = x;
if(x<0) {
tmp = -x;
}
while(tmp>0) {
result=result*10+(tmp%10);
tmp=tmp/10;
}
if(result>Integer.MAX_VALUE || result<Integer.MIN_VALUE) {
return 0;
}
return x<0? -(int)result:(int)result;
}