Open micfan opened 9 years ago
Thanks for pointing this out, I'll check it later.
signed long long int r=0;
for(;x;x/=10)
r = r*10+x%10;
if(abs(r)>(pow(2,31)-1)) return 0;
else return (int) r;
@billlipeng Thanks! https://oj.leetcode.com/problems/reverse-integer/
#include <math>
class Solution {
public:
int reverse(int x) {
signed long long int result = 0;
for (; x; x/=10)
{
result = result * 10 + x % 10;
}
if (abs(result) > (pow(2, 31)-1)) {
return 0;
} else {
return result;
}
}
};
+1. The solution is wrong.
谢谢,我周末抽时间检查一下
This issue was solved. Plz check my comment on Feb 26, update the code and close it. Thanks a lot!
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?
yeah?