To estimate the root of f(x), in each iteration, we update x with :

x(j+1) = x(j) - f(x(j)) / f'(x(j))

In this problem, f(x) = x^2 - a

a represents the original number which we want to find the square root.

```
public int mySqrt(int x) {
double y = 600.1;
for(int i = 0; i < 10; ++i)
y = 0.5 * y + x * 0.5 / y;
return (int)y;
}
```