Reverse Integer
Question
Given a 32-bit signed integer, reverse digits of an integer.
Example 1:
Input: 123
Output: 321
Example 2:
Input: -123
Output: -321
Example 3:
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.
Answer
问题不算难，设置一个while然后求余并除应该可以得到结果
class Solution {
public:
int reverse(int x) {
int temp = x;
int result = 0;
while(temp) {
result = result*10 + temp % 10;
temp /= 10;
}
return result;
}
};
但是题目有一个要求：机器是运行在32位的系统上的，溢出时需要改为0。
因此需要修改。
首先我们将结果改为long long的类型，再去判断结果是否超过32位的最值
long long result = 0;
...
return (result > INT_MAX || result < INT_MIN) ? 0 : result;
