Problem
Given a signed 32-bit integer x, return x with its digits reversed. If reversing the digits causes the result to fall outside the signed 32-bit integer range, return0.
The 32-bit signed integer range is from
-231 to 231 - 1.
Example(s)
Consider the following examples to understand the expected input and output.Example 1
Input
x = 123Output
321Example 2
Input
x = -123Output
-321Solution
This solution uses the Math / Number Theory pattern. We extract the last digit usingx % 10 and add it to reversed after shifting the existing digits one position to the left.
We use a long variable for the reversed number so that we can safely detect whether the result exceeds the 32-bit integer range before converting it back to
int.
class Solution {
public int reverse(int x) {
long reversed = 0;
while (x != 0) {
int a = x % 10;
// Add the last digit to the reversed number.
reversed = reversed * 10 + a;
// Remove the last digit from x.
x = x / 10;
}
// Check if the reversed number exceeds the integer range.
if (reversed > Integer.MAX_VALUE || reversed < Integer.MIN_VALUE) {
return 0;
}
return (int) reversed;
}
}Complexity
We process each digit of the number once, so the time complexity isO(log x), where the number of digits determines the number of iterations.
The solution uses only a few variables, so the extra space complexity is
O(1).