Given a non-negative integer
x, compute and return the square root of
Since the return type is an integer, the decimal digits are truncated, and only the integer part of the result is returned.
Input: x = 4 Output: 2
Input: x = 8 Output: 2 Explanation: The square root of 8 is 2.82842..., and since the decimal part is truncated, 2 is returned.
0 <= x <= 231 - 1
Solution in Python
class Solution: def mySqrt(self, x: int) -> int: return int(x**0.5)