[Leetcode brushing notes] 69. The square root of x

69. Square root of x

topic description

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 kept, and the decimal part will be discarded.

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

input Output

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

Newton iteration method

Equivalent to solving $f(x)=x^2-$positive root of$a$

use $x_n$The tangent to intercept the x-axis, get the abscissa $x_{n+1}$, continue with $x_{n+1}$To cut the x-axis, it will gradually converge to the root of the function

the code

class Solution {
public:
    int mySqrt(int x) {
        long res = x;
        while (res * res > x)
            res = (res + x / res) / 2;
        return (int) res;
    }
};