LEETCODE--SQRT (x)

Description:

Implement int sqrt(int x) .

Compute and return the square root of X.

It is very useful to study maths well, and Newton's iterative method solves it.

computes the^{solution of x 2 = n, making f (x) =x2-n equivalent to solving the solution of the F (x) =0, as shown in the figure on the left.}

First Take x_{0, if x0 is not a solution, make atangent to the point (X 0,f (x0)), and the x-axis intersection is x1.}

Similarly, if X_{1 is not a solution, make atangent of this point (x 1,f (x1)), and the x-axis intersection is x2.}

And so on

The x I obtained in this way_{is infinitely nearer to the solution of f (x) =0.}

There are_{two ways to determine if x i is an F (x) =0 solution:}

First_{, the direct calculation of the value of f (x i) is 0, and the second is to judge whether the two solutions xI and xi-1 are infinitely close.}

public class Solution {public    int mysqrt (int x) {        if (x ==0)             return 0;          Double pre;          Double cur = 1;          Do          {              pre = cur;              cur = x/(2 * pre) + pre/2.0;          } while (Math.Abs (cur-pre) > 0.00001);          return (int) cur;}              }

LEETCODE--SQRT (x)