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7. Reverse Integer

  • Time: $O(\log x)$
  • Space: $O(1)$
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class Solution {
 public:
  int reverse(int x) {
    long ans = 0;

    while (x != 0) {
      ans = ans * 10 + x % 10;
      x /= 10;
    }

    return (ans < INT_MIN || ans > INT_MAX) ? 0 : ans;
  }
};

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class Solution {
  public int reverse(int x) {
    long ans = 0;

    while (x != 0) {
      ans = ans * 10 + x % 10;
      x /= 10;
    }

    return (ans < Integer.MIN_VALUE || ans > Integer.MAX_VALUE) ? 0 : (int) ans;
  }
}

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class Solution:
  def reverse(self, x: int) -> int:
    ans = 0
    sign = -1 if x < 0 else 1
    x *= sign

    while x:
      ans = ans * 10 + x % 10
      x //= 10

    return 0 if ans < -2**31 or ans > 2**31 - 1 else sign * ans

var reverse = function(x) {
  let rev = 0;
  while (x !== 0) {
    rev = rev * 10 + (x % 10);
    x = Math.trunc(x / 10);
  }
  if (rev < -(2 ** 31) || rev > 2 ** 31 - 1) return 0;
  return rev;
};
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