README_EN.md

7. Reverse Integer

中文文档

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

Approach 1: Pop and Push Digits & Check before Overflow

Time complexity $O(log|x|)$, Space complexity $O(1)$.

Python3

class Solution:
    def reverse(self, x: int) -> int:
        y = int(str(abs(x))[::-1])
        res = -y if x < 0 else y
        return 0 if res < -(2**31) or res > 2**31 - 1 else res

Java

class Solution {
    public int reverse(int x) {
        long res = 0;
        while (x != 0) {
            res = res * 10 + (x % 10);
            x /= 10;
        }
        return res < Integer.MIN_VALUE || res > Integer.MAX_VALUE ? 0 : (int) res;
    }
}

C++

class Solution {
public:
    int reverse(int x) {
        int ans = 0;
        for (; x != 0; x /= 10) {
            if (ans > INT32_MAX / 10 || ans < INT32_MIN / 10)
                return 0;
            ans = ans * 10 + x % 10;
        }
        return ans;
    }
};

JavaScript

/**
 * @param {number} x
 * @return {number}
 */
var reverse = function (x) {
    let res = 0;
    while (x) {
        res = res * 10 + (x % 10);
        x = ~~(x / 10);
    }
    return res < Math.pow(-2, 31) || res > Math.pow(2, 31) - 1 ? 0 : res;
};

C

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

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,
        }
    }
}

Go

func reverse(x int) int {
	ans, INT32_MAX, INT32_MIN := 0, math.MaxInt32, math.MinInt32
	for ; x != 0; x /= 10 {
		if ans > INT32_MAX/10 || ans < INT32_MIN/10 {
			return 0
		}
		ans = ans*10 + x % 10
	}
	return ans
}

C#

public class Solution {
    public int Reverse(int x) {
        var negative = x < 0;
        if (negative) x = -x;
        long result = 0;
        while (x > 0)
        {
            result = (result * 10) + x % 10;
            x /= 10;
        }
        if (negative) result = -result;
        if (result > int.MaxValue || result < int.MinValue) result = 0;
        return (int) result;
    }
}

Ruby

# @param {Integer} x
# @return {Integer}
def reverse(x)
  neg = x < 0

  x = x.abs
  s = ''

  x /= 10 while x > 0 && (x % 10).zero?

  while x > 0
    s += (x % 10).to_s
    x /= 10
  end

  s = neg ? '-' + s : s

  # have to explicitly constraint the int boundary as per the dummy test case
  res = s.to_i
  res <= 214_748_364_7 && res >= -214_748_364_8 ? res : 0
end

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