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https://github.com/SunnyQjm/algorithm-review.git
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77 lines
2.5 KiB
Python
77 lines
2.5 KiB
Python
#!/usr/bin/env python
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# coding=utf-8
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################################################################
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# LeetCode: 69 Sqrt(x)
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# Implement int sqrt(int x).
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# Compute and return the square root of x, where x is guaranteedto be a non-negative integer.
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# Since the return type is an integer, the decimal digits aretruncated and only the integer part of the result is returned.
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################################################################
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import math
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class Solution:
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def mySqrt(self, x):
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"""
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(Knowledge)
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函数功能描述: 传入一个数字，返回其平方根的整数部分
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思路：
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1. 首先用特判，处理掉x=1的特殊情况；
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2. 接着用二分法，找到x平方根的整数部分；
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二分法的结束条件：跳出循环之前的最后一次循环，left == right, 此时得到 mid == left，因此：
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- 如果mid > x * mid，则 x 的平方根必然为比left略小一点的某个值，所以最后结果返回 left - 1
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- 如果mid <= x * mid, 则 x 的平方根必然为比left略大一点某个值，又因为执行了 left = mid +１, 所以最后结果返回 left - 1
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"""
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if x < 2:
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return x
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left, right = 1, x // 2
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while left <= right:
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mid = left + (right - left) // 2
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if mid > x / mid:
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right = mid - 1
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else:
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left = mid + 1
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return left - 1
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def mySqrt2(self, x):
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"""
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(Knowledge)
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算法说明：使用牛顿迭代法，近似求平方根
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牛顿迭代法求平方跟的证明详见 => https://www.cnblogs.com/upcan/p/9907402.html
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最终得到迭代公式： next = 1/2 * (next + input / next)
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- next => 等式左边为下一个近似值，等式右的next表示上一个近似值
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- input => 输入值
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"""
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# 首先猜测一个初始的预测值
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next = x / 2
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# 运行20次迭代（这个迭代次数取决于需要的精度，迭代次数越多，得到的结果越精确）
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for k in range(20):
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next = 1 / 2 * (next + x / next)
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# 根据题意，进行向下取整
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return math.floor(next)
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if __name__ == '__main__':
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solution = Solution()
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print(solution.mySqrt(9))
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print(solution.mySqrt(8))
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print(solution.mySqrt(10))
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print(solution.mySqrt(7))
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print(solution.mySqrt2(9))
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print(solution.mySqrt2(8))
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print(solution.mySqrt2(10))
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print(solution.mySqrt2(7))
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