README_EN.md

7. Reverse Integer

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

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

• When $ans<\lfloor \frac{mx}{10}\rfloor$, the inequality obviously holds;
• When $ans=\lfloor \frac{mx}{10}\rfloor$, the necessary and sufficient condition for the inequality to hold is $y\le 7$. If $ans=\lfloor \frac{mx}{10}\rfloor$ 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 $ans>\lfloor \frac{mx}{10}\rfloor$, the inequality obviously does not hold.

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

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

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

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

Python3

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

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++

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

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
}

JavaScript

/**
 * @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

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;
}

Rust

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,
        }
    }
}

C#

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;
    }
}

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