7. Reverse Integer

Description

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

Solutions

Solution 1: Mathematics

Let’s denote $mi$ and $mx$ as $-2^{31}$ and $2^{31} - 1$ respectively, then the reverse result of $x$, $ans$, needs to satisfy $mi \le ans \le mx$.

We can continuously take the remainder of $x$ to get the last digit $y$ of $x$, and add $y$ to the end of $ans$. Before adding $y$, we need to check if $ans$ overflows. That is, check whether $ans \times 10 + y$ is within the range $[mi, mx]$.

If $x \gt 0$, it needs to satisfy $ans \times 10 + y \leq mx$, that is, $ans \times 10 + y \leq \left \lfloor \frac{mx}{10} \right \rfloor \times 10 + 7$. Rearranging gives $(ans - \left \lfloor \frac{mx}{10} \right \rfloor) \times 10 \leq 7 - y$.

Next, we discuss the conditions for the inequality to hold:

When $ans \lt \left \lfloor \frac{mx}{10} \right \rfloor$, the inequality obviously holds;
When $ans = \left \lfloor \frac{mx}{10} \right \rfloor$, the necessary and sufficient condition for the inequality to hold is $y \leq 7$. If $ans = \left \lfloor \frac{mx}{10} \right \rfloor$ and we can still add numbers, it means that the number is at the highest digit, that is, $y$ must not exceed $2$, therefore, the inequality must hold;
When $ans \gt \left \lfloor \frac{mx}{10} \right \rfloor$, the inequality obviously does not hold.

In summary, when $x \gt 0$, the necessary and sufficient condition for the inequality to hold is $ans \leq \left \lfloor \frac{mx}{10} \right \rfloor$.

Similarly, when $x \lt 0$, the necessary and sufficient condition for the inequality to hold is $ans \geq \left \lfloor \frac{mi}{10} \right \rfloor$.

Therefore, we can check whether $ans$ overflows by checking whether $ans$ is within the range $[\left \lfloor \frac{mi}{10} \right \rfloor, \left \lfloor \frac{mx}{10} \right \rfloor]$. If it overflows, return $0$. Otherwise, add $y$ to the end of $ans$, and then remove the last digit of $x$, that is, $x \gets \left \lfloor \frac{x}{10} \right \rfloor$.

The time complexity is $O(\log |x|)$, where $|x|$ is the absolute value of $x$. The space complexity is $O(1)$.

Python Code

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class Solution:
    def reverse(self, x: int) -> int:
        ans = 0
        mi, mx = -(2**31), 2**31 - 1
        while x:
            if ans < mi // 10 + 1 or ans > mx // 10:
                return 0
            y = x % 10
            if x < 0 and y > 0:
                y -= 10
            ans = ans * 10 + y
            x = (x - y) // 10
        return ans

Java Code

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class Solution {
    public int reverse(int x) {
        int ans = 0;
        for (; x != 0; x /= 10) {
            if (ans < Integer.MIN_VALUE / 10 || ans > Integer.MAX_VALUE / 10) {
                return 0;
            }
            ans = ans * 10 + x % 10;
        }
        return ans;
    }
}

C++ Code

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class Solution {
public:
    int reverse(int x) {
        int ans = 0;
        for (; x; x /= 10) {
            if (ans < INT_MIN / 10 || ans > INT_MAX / 10) {
                return 0;
            }
            ans = ans * 10 + x % 10;
        }
        return ans;
    }
};

Go Code

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func reverse(x int) (ans int) {
	for ; x != 0; x /= 10 {
		if ans < math.MinInt32/10 || ans > math.MaxInt32/10 {
			return 0
		}
		ans = ans*10 + x%10
	}
	return
}

Rust Code

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impl Solution {
    pub fn reverse(mut x: i32) -> i32 {
        let is_minus = x < 0;
        match x.abs().to_string().chars().rev().collect::<String>().parse::<i32>() {
            Ok(x) => x * (if is_minus { -1 } else { 1 }),
            Err(_) => 0,
        }
    }
}

JavaScript Code

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/**
 * @param {number} x
 * @return {number}
 */
var reverse = function (x) {
    const mi = -(2 ** 31);
    const mx = 2 ** 31 - 1;
    let ans = 0;
    for (; x != 0; x = ~~(x / 10)) {
        if (ans < ~~(mi / 10) || ans > ~~(mx / 10)) {
            return 0;
        }
        ans = ans * 10 + (x % 10);
    }
    return ans;
};

C# Code

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public class Solution {
    public int Reverse(int x) {
        int ans = 0;
        for (; x != 0; x /= 10) {
            if (ans < int.MinValue / 10 || ans > int.MaxValue / 10) {
                return 0;
            }
            ans = ans * 10 + x % 10;
        }
        return ans;
    }
}

C Code

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int reverse(int x) {
    int ans = 0;
    for (; x != 0; x /= 10) {
        if (ans > INT_MAX / 10 || ans < INT_MIN / 10) {
            return 0;
        }
        ans = ans * 10 + x % 10;
    }
    return ans;
}

PHP Code

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class Solution {
    /**
     * @param int $x
     * @return int
     */

    function reverse($x) {
        $isNegative = $x < 0;
        $x = abs($x);

        $reversed = 0;

        while ($x > 0) {
            $reversed = $reversed * 10 + ($x % 10);
            $x = (int) ($x / 10);
        }

        if ($isNegative) {
            $reversed *= -1;
        }
        if ($reversed < -pow(2, 31) || $reversed > pow(2, 31) - 1) {
            return 0;
        }

        return $reversed;
    }
}

7. Reverse Integer#

Description#

Solutions#

Solution 1: Mathematics#

7. Reverse Integer

Description

Solutions

Solution 1: Mathematics