Leetcode/python/69_sqrtx.py
2024-04-08 16:56:16 +01:00

13 lines
369 B
Python

class Solution:
def mySqrt(self, x: int) -> int:
# Use Newton-Raphson iteration
# y**2 - x = 0 = f(y)
# f'(y) = 2y
# y = y - (y**2 - x)/(2y)
y = 1
close_enough = False
while not close_enough:
y -= (y*y-x)/(2*y)
close_enough = y*y - x <= 0.1 and (y+1)*(y+1) > x
return floor(y)