[Leetcode] SQRT (x)

SQRT (x)

Implement int sqrt(int x) .

Compute and return the square root of X.

Problem Solving Ideas:

The problem is to find the square root of x.

Solution 1: The basic idea is to enumerate, from 1 to N to traverse until result*result>x, then the result is result-1.

Class Solution {public:    int mysqrt (int x) {        int result = 0;        while (long long) result * result <= x) {            result++;        }        return result-1;}    ;

However, this creates a timeout error.

Solution 2: Using a method similar to binary search, the results can be found by binary search in the 0-x interval. Notice that the multiplication may overflow.

Class Solution {public:    int mysqrt (int ×) {        long long left = 0, right = x;        Long Long middle = (left+right)/2;        while (left<right) {            long long temp = Middle*middle;            if (temp = = x) {                return middle;            } else if (temp > x) {right                = middle-1;            } else{Left                = middle + 1;            }            Middle = (left + right)/2;        }        if (Long Long) middle*middle> (long Long) x) {            return (int) middle-1;        } else{            return (int) middle;}}    ;

Solution 3: Newton iterative Method (Reference blog: http://www.cnblogs.com/AnnieKim/archive/2013/04/18/3028607.html). Here is the citation.

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-xi", 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.
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With the iterative formula, the program is ready to write. About Newton's iterative method, you can refer to Wikipedia and Baidu Encyclopedia .

Newton's Iterative method can also be used to solve the multiple-equation solution.

P.S. The problem is to solve the square root of an integer, and the return value is an integral type. A slight modification based on the above code can also be applied to double (method 2 only).

Double sqrt (double x) {    if (x = = 0) return 0;    Double last = 0.0;    Double res = 1.0;    while (res! =)    {last        = res;        Res = (res + x/res)/2;    }    return res;}

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[Leetcode] SQRT (x)