Viiisit [LeetCode] - 69. Sqrt(x)

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Problem - Sqrt(x)

Given a non-negative integer x, return the square root of x rounded down to the nearest integer. The returned integer should be non-negative as well.

You must not use any built-in exponent function or operator.

For example, do not use pow(x, 0.5) in c++ or x ** 0.5 in python.

翻譯蒟蒻

給定一個正整數，找出不大於 x 的最大平方值所對應的整數。（不能用內建函數完成）

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• Example 1:

  Input: x = 4
  Output: 2
  Explanation: The square root of 4 is 2, so we return 2.

• Example 2:

  Input: x = 8
  Output: 2
  Explanation: The square root of 8 is 2.82842..., and since we round it down to the nearest integer, 2 is returned.

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Solution - JavaScript

/**
 * @param {number} x
 * @return {number}
 */
var mySqrt = function(x) {
  let tmp = x;
  while (tmp * tmp > x) {
    tmp = Math.floor((tmp + x / tmp) / 2);
  }
  return tmp;
};

Solution - Ruby

def my_sqrt(x)
  tmp = x
  while tmp * tmp > x
    tmp = (tmp + x / tmp) / 2
  end
  tmp 
end

Solution - PHP

function mySqrt($x) {
  $tmp = $x;
  while ($tmp * $tmp > $x) {
    $tmp = intval(($tmp + $x / $tmp) / 2);
  }
  return $tmp;
}

LeetCode 傳送門 - Sqrt(x)