7. Reverse Integer

Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes the value to go outside the signed 32-bit integer range [-231, 231 - 1], then return 0.

Assume the environment does not allow you to store 64-bit integers (signed or unsigned).

 

Example 1:

Input: x = 123
Output: 321

Example 2:

Input: x = -123
Output: -321

Example 3:

Input: x = 120
Output: 21

 

Constraints:

  • -231 <= x <= 231 - 1

Approach 01: Math Pop/Push



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;
    }
};

Time Complexity:

  • O(log(x)): There are roughly log10(x) digits in x.

Space Complexity:

  • O(1): Constant space used.
class Solution:
    def reverse(self, x: int) -> int:
        rev = 0
        sign = 1 if x >= 0 else -1
        x *= sign
        while x != 0:
            pop = x % 10
            x //= 10
            if rev > (2**31 - 1) // 10 or (rev == (2**31 - 1) // 10 and pop > 7):
                return 0
            rev = rev * 10 + pop
        return rev * sign

Time Complexity:

  • O(log(x)): There are roughly log10(x) digits in x.

Space Complexity:

  • O(1): Constant space used.
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;
    }
}

Time Complexity:

  • O(log(x)): There are roughly log10(x) digits in x.

Space Complexity:

  • O(1): Constant space used.

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