1.Reverse digits of an integer.
Example1: x = 123, return 321
Example2: x = -123, return -321
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?
For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
2.Note:
The input is assumed to be a 32-bit signed integer. Your function should return 0 when the reversed integer overflows.
3.題目要求：翻轉(zhuǎn)數(shù)字邑蒋，需要考慮溢出問(wèn)題∫嬉В可以用long long型變量保存計(jì)算結(jié)果，最后返回的時(shí)候判斷是否在int返回內(nèi)斧拍。
4.代碼：
class Solution {
public:
int reverse(int x) {
long long res = 0;
while (x != 0) {
res = 10 * res + x % 10;
x /= 10;
}
return (res > INT_MAX || res < INT_MIN) ? 0 : res;
}
};
