Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
Have you thought about this?
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).
class Solution {
public:
int reverse(int x)
{
long result = 0;
int sign = x > 0 ? 1 : -1;
x *= sign;
int mod = 0;
while (x > 0) {
mod = x % 10;
result = result * 10 + mod;
x /= 10;
}
if (result > 2147483647) {
return 0;
}
return sign * (int)result;
}
};
本文探讨了如何实现整数反转并处理边界情况,包括最后一位为0的整数和可能的溢出问题。
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