69. Sqrt(x)

https://leetcode.com/problems/sqrtx/description/

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.
 

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.
 

Constraints:

0 <= x <= 231 - 1

응? 뭐지?
걍 제곱근 구해서 내림하면 되는데...?

class Solution:
    def mySqrt(self, x: int) -> int:
        ans = math.floor(math.sqrt(x))
        return ans