leetcode——69.x的平方根

1 class Solution:
2     def mySqrt(self, x: int) -> int:
3         for i in range(0,x//2+2):
4             if x>=i**2 and x<(i+1)**2:
5                 return i

 1 class Solution:
 2     def mySqrt(self, x: int) -> int:
 3         i=x
 4         m=[x]
 5         while i>=0:
 6             if i**2>x:
 7                 m[0]=i
 8                 i=i//2
 9             elif x>=i**2 and x<(i+1)**2:
10                 return i
11             else:
12                 i=(i+m[0])//2