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Portal ——>Leecode Big factory hot topic 100 Tao series problem solution
Problem description
To give you one 32 Signed integer of bit x , Return to x The result of reversing the number part in .
If the integer after inversion exceeds 32 The range of signed integers of bits [−231, 231 − 1] , Just go back to 0.
Suppose the environment doesn't allow storage 64 An integer （ With or without sign ）.
Example 1：
Input ：x = 123
Output ：321
Example 2：
Input ：x = -123
Output ：-321
Example 3：
Input ：x = 120
Output ：21
Example 4：
Input ：x = 0
Output ：0
Tips ：
-231 <= x <= 231 - 1
Attached is the complete runnable code
#include "iostream"
#include "algorithm"
#include "cmath"
using namespace std;
class Solution {
public:
int reverse(int x) {
bool flag = false;
if (x > 0) flag = !flag;
long long res = 0;
while(x) {
res = res * 10 + (x % 10);
x /= 10;
}
if(res < -pow(2, 31) || res > pow(2, 31) - 1) return 0;
return flag ? abs(res) : -abs(res);
}
};
int main() {
Solution solution;
cout << solution.reverse(1534236469);
return 0;
}