Description
Given a 32-bit signed integer, reverse digits of an integer.
Input: 123
Output: 321
Input: -123
Output: -321
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.
Solution
reverse一个整数，考虑到负数和溢出就行了
int reverse(int x) {
long long n = 0;
while (x) {
n = n * 10 + x % 10;
x /= 10;
}
return (n > INT_MAX || n < INT_MIN)? 0 : n;
}
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