141. SQRT (x) "Newton iterative method for square root by Java"

Description

Implement int sqrt(int x) .

Compute and return the square root of X.

Example

sqrt (3) = 1

sqrt (4) = 2

sqrt (5) = 2

sqrt (10) = 3

Challenge

O (log (x))

Test instructions: To find the square root of a given number, if you use a general method, such as a dichotomy, you need to consider the scope of the int type, do not overflow. The best method when Newton iterative method. The code is as follows:

 Public classSolution {/**     * @paramX:an Integer *@return: The sqrt of x*/     Public intsqrtintx) {//Write your code here//Newton's iterative method to find square root       Doublep=2.0; DoublePre=0;  while(Math.Abs (P-pre) >1e-6) {Pre=p; P= (p+x/p)/2.0; }       return(int) p; }}

On the application principle of Newton iterative method on square root, see blog: http://www.matrix67.com/blog/archives/361

141. SQRT (x) "Newton iterative method for square root by Java"