题目:
Implement int sqrt(int x).
Compute and return the square root of x, where x is guaranteed to be a non-negative integer.
Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.
Example 1:
Input: 4
Output: 2
Example 2:
Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since
the decimal part is truncated, 2 is returned.
思路:
牛顿迭代法, 导数方程 f'(x)*x'=y', 任何函数f(x)=y,求解某个y=n,均可以转化为 f(x)-n=0,
此后就可以用牛顿迭代法,不断逼近实际待求x值。
牛顿迭代共识:f'(x_pre)x_pre+x_pre=x_cur
应用: 迭代思想,类似于 动态规划思想,pre==>cur,进行动态推断处理
class Solution:
def mySqrt(self, x):
"""
:type x: int
:rtype: int
"""
r=x/2+1
while r*r-x>1e-10:
r=(r+x/r)/2
# print(r*r-x)
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