Reverse Integer(翻轉整數)
阿新 • • 發佈:2019-02-15
Have you thought about this?
if (x < 0)
{
flag = -1;
x = -x;
}
int num = 0;
while (x > 0)
{
//if ((num != 0 && (INT_MAX / abs(num) < 10)) || ((unsigned int)abs(num * 10) + (unsigned int)(x % 10) > INT_MAX))
//{//溢位
//return 0;
//}
num = num * 10 + flag * (x % 10);
x = x / 10;
}
return num;
}
};
Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
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?
Throw an exception? Good, but what if throwing an exception is not an option? You would then have to re-design the function (ie, add an extra parameter).
實現程式碼:
class Solution {
public:
int reverse(int x) {
- #define INT_MAX 2147483647
- #define INT_MIN (-INT_MAX - 1)
if (x < 0)
{
flag = -1;
x = -x;
}
int num = 0;
while (x > 0)
{
//if ((num != 0 && (INT_MAX / abs(num) < 10)) || ((unsigned int)abs(num * 10) + (unsigned int)(x % 10) > INT_MAX))
//{//溢位
//return 0;
//}
num = num * 10 + flag * (x % 10);
x = x / 10;
}
return num;
}
};