Question

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 $x \gt 0$, 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 $O(1)$.

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

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;
  }
}

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;
  }
};

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
}

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,
    }
  }
}

/**
 * @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;
};

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;
  }
}

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;
}

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;
  }
}