leetcode--Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

public class Solution {
    public int sqrt(int x) {
       if(x < 0) return -1;
		if(x == 0) return 0;
		
		//binary search method
		long high = x / 2 + 1;
		long low = 0;
		while(low <= high) {
			long mid = low + (high - low) / 2;
			long approximateSquare = mid * mid;
			if(approximateSquare == x)
				return (int) mid;
			
			if(approximateSquare < x) //too small
				low = mid + 1;
			else //too large
				high = mid - 1;
		}
		return (int) high;
    }
}

newton iteration method (x_{i+1} = x_i - f(x_i) / f^'(x_i)). where f^'(x_i) = 2x_i in this problem.

Therefore, x_{i+1} = x_i -(x_i^2 - x) /2x_i = ( x_i + x / x_i) / 2. The algorithm converges to the root, so the termination condition of the loop is x_{i} == x{i+1}

public class Solution {
    public int sqrt(int x) {
        if(x < 0) return -1;
		if(x == 0) return 0;
		
		double last = 0;
		double iter = 1;
		while(last != iter){ //termination condition. the algorithm converges to the root
			last = iter;
			iter = (iter + x / iter) / 2;
		}
		return (int) iter;
    }
}