LeetCode ❀ 69. Square root of x / python pocket calculator algorithm to find the square root

topic:

Given a non-negative integer x, compute and return the arithmetic square root of x .

Since the return type is an integer, only the integer part of the result is retained , and the decimal part will be rounded off .

Note : Any built-in exponential functions and operators are not allowed, such as pow(x, 0.5) or x ** 0.5.

Example 1:

输入：x = 4
输出：2

 Example 2:

输入：x = 8
输出：2
解释：8 的算术平方根是 2.82842..., 由于返回类型是整数，小数部分将被舍去。

answer:

Instead of arithmetic square roots, we use  the pocket calculator algorithm

class Solution:
    def mySqrt(self, x: int) -> int:
        if x == 0:return 0
        ans = int(math.exp(0.5 * math.log(x)))
        return ans + 1 if (ans + 1) ** 2 <= x else ans