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.
解决办法:
class Solution {
public int reverse(int x) {
int result=0;
while(x!=0){
int tail=x%10;
int newResult=result*10+tail;
if((newResult-tail)/10!=result){
return 0;
}
x=x/10;
result=newResult;
}
return result;
}
}
本来这是一个简单的翻转数字,但是给的是32位的整数,因此需要有所判断,超出界限就应该return 0
那么该如何判断呢?假设得出的newResult是溢出的,那么它得出的值是混乱的,因此利用newResult重新逆着计算就会得出不一样的答案。