Leetcode | | 69, SQRT (x)

Problem:

Implement int sqrt(int x) .

Compute and return the square root of X.

Hide TagsMath Binary SearchTest instructions: Calculates square root, no precision specified, default precision is 0.00001

Thinking:

(1) Because the problem does not set the accuracy, suggesting the use of binary method

(2) In addition to the dichotomy, you can also use Newton's iterative

Code

Two-part method:

Class solution{public    :    int sqrt (int x) {        unsigned long long begin = 0;        unsigned long long end = (x+1)/2;        unsigned long long mid;        unsigned long long tmp;        while (begin < End)        {            mid = begin + (End-begin)/2;            TMP = Mid*mid;            if (tmp==x) return mid;            else if (tmp<x) begin = mid+1;            else end = Mid-1;        }        TMP = End*end;        if (tmp > x)            return end-1;        else            return end;}    };

Newton Iteration:

To facilitate understanding, first take the subject as an example:

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.}

after (x_{i, F (x i) the tangent equation for this point is f (x) = f (xi) + F ' (Xi) (x-x< Sub style= "margin:0px; padding:0px ">i", where F ' (x) is the derivative of f (x), 2x in the subject. So that the tangent equation equals 0, you can find Xi+1=xi -F (xi"/F ' (Xi).}

Continue to simplify, x_{i+1=x i -(xi^{2 -n"/(2xi) = Xi  -xi /2 + N/(2xi) = xi /2 + n/2xi = (x i + n/xi)/2.
/sub>}}

With the iterative formula x_{i+1= (xi + n/xi)/2}, the program is ready to write. About Newton's iterative method, you can refer to Wikipedia and Baidu Encyclopedia .

Class Solution {public:    int sqrt (int x) {        if (x<0)         return-1;        Double a=1.0;        Double check=0;        do{            a= (x/a+a)/2;            check = A*a;        } while (ABS (CHECK-X) >0.00001);        return A;    };

Leetcode | | 69, SQRT (x)