Sqrt(x) LeetCode

Question:

https://oj.leetcode.com/problems/sqrtx/

解答：

1）二分法。

取中点，如果当前点平方大于x，前半段继续二分。否则后半段继续二分。需要注意的是求的值是平方最接近但是小于，或者等于x。所以中点如果小于x，后半段需要包括这个中点。这就需要中点能够大于开始的点。其次还有一点需要注意的是平方可能超过int的最大值，简单的做法就是用long long来保存这个平方。

    int sqrt2(int x) {
        
        if (x <= 0)
            return 0;
        
        if (x == 1)
            return 1;
    
        int s = 1;
        int e = x/2;
        
        while (1)
        {
            if (s == e)
                return s;
            
            // make sure middle is always > s    
            int middle = (s+e)/2+1;
            
            long long m2 = (long long)middle*(long long)middle;
            
            if (m2 == x)
            {
                return middle;
            }
            else if (m2 > x)
            {
                e = middle-1;
            }
            else
            {
                s = middle;
            }
        }
        
        return 0;
    }

 

2）牛顿迭代法

这是数学上的求近似值的办法。

From Wiki：

In numerical analysis, Newton's method (also known as the Newton–Raphson method), named after Isaac Newton and Joseph Raphson, is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function.

The Newton–Raphson method in one variable is implemented as follows:

Given a function ƒ defined over the reals x, and its derivative ƒ', we begin with a first guess x₀ for a root of the function f. Provided the function satisfies all the assumptions made in the derivation of the formula, a better approximation x₁ is

Geometrically, (x₁, 0) is the intersection with the x-axis of the tangent to the graph of f at (x₀, f (x₀)).

The process is repeated as

until a sufficiently accurate value is reached.

 

对于函数f(x)=0，和任意一个起点x_i，持续一下的步骤直到x_i和x_i+1足够相同。

x_i+1=x_i-f(x)/f'(x). 

对于我们这个问题，f(x) = x²-n，所以f'(x) = 2x.

x_i+1=x_i-f(x)/f'(x) = xi-(x_i²-n)/(2x_i) = (x_i+n/x_i)/2

因为最终的近似结果平方可能大于x也可能小于x（不太确定，因为最终结果是double取整，本身已经小于value），所以保险起见需要最后判断需要不需要减一。

    int sqrt(int x)
    {
        const double dmin = 0.000001;
        
        double value = 1;
        double last = 0;
        
        while (abs(value - last) > dmin)
        {
            last = value;
            value = (value + x/value) / 2;
        }
        
        int n = (int)value;
        
        if (n*n > x)
            n = n-1;
            
        return n;
    }