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 store integers within the 32-bit signed integer range: [−231, 231 − 1]. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
Solution 1 in C++
#include <iostream>
#include <math.h>
using namespace std;
class Solution {
public:
int reverse(int x) {
int y=x;
long ret=0;
long valid_range = pow(2,31);
while(y)
{
ret *= 10;
ret += (y %10);
cout << ret << " - " << y << endl;
y /= 10;
}
if ((ret<valid_range*(-1)) || (ret>(valid_range-1))) return 0;
return ret;
}
};
int main(void)
{
Solution s;
int testcase;
testcase = 123;
cout << testcase << " -> " << s.reverse(testcase) << endl;
testcase = -123;
cout << testcase << " -> " << s.reverse(testcase) << endl;
testcase = 120;
cout << testcase << " -> " << s.reverse(testcase) << endl;
testcase = 1534236469;
cout << testcase << " -> " << s.reverse(testcase) << endl;
return 0;
}
Solution 2 in C++
class Solution {
public:
int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > INT_MAX/10 || (rev == INT_MAX / 10 && pop > 7)) return 0;
if (rev < INT_MIN/10 || (rev == INT_MIN / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
};
Solution 3 in Java
class Solution {
public int reverse(int x) {
int rev = 0;
while (x != 0) {
int pop = x % 10;
x /= 10;
if (rev > Integer.MAX_VALUE/10 || (rev == Integer.MAX_VALUE / 10 && pop > 7)) return 0;
if (rev < Integer.MIN_VALUE/10 || (rev == Integer.MIN_VALUE / 10 && pop < -8)) return 0;
rev = rev * 10 + pop;
}
return rev;
}
}