LeetCode69 Sqrt (x)

Link Address: https://leetcode.com/problems/sqrtx/

The problem is to ask for the square root of a number.

I'm here to offer three ways

1: Everyone knows that the square root must be between [1,X/2], so from 1 cycles to X/2, but when X=1 is passed, not a good method and will tle

Class Solution {    //tle and imprecise public:    int sqrt (int x) {        int t = X/2;        for (int i = 0; i<= t; i++)            if ((i * i) = = x) return i;        return-1;    }};

2: Dichotomy

We know that the time complexity required for the dichotomy is O (LgN), so that it does not time out.

Class Solution {public:    int sqrt (int x) {        double begin = 0;        Double end = x;        double result = 1;        Double mid = 1;        while (ABS (result-x) > 0.000001) {            mid = (begin+end)/2;            result = Mid*mid;            if (Result > x)   //Two-point range                end = mid;            else begin = Mid;        }        return (int) mid;    }};

3: Newton Iterative method

NewtonIterative method(Newton ' s method) also known asNewton-Raphson (Kristinn) methods (Newton-raphson method), it isNewtonpresented in 17th century as aRealDomain andPluralon the domainApproximate solution equation the method. set R yes

the root, selectas the initial approximation of R, over PointdoCurvethe equation for the tangent l,l isto find the horizontal axis of the intersection of L and X axis, called X 1is an approximate value of R. Over DotMake a curvetangent and the horizontal axis of the x-axis intersection, saidtwo approximate values for R. Repeat the above process to get a sequence of approximate values of R, where,called R.The second approximation, the above-namedNewton's iterative formula.

The following code cur = pre/2 + x/(2*pre) is the result of the simplification calculation. Here's f (x) = X^2-n

* Newton Iterative Method */class Solution {public:    int sqrt (int x) {        double pre = 0;        Double cur = x;           Starting at x starting from  X/2 will cause 1 to not meet  x (n+1) = Xn-f ' (xn)/f (xn) while         (ABS (CUR-PRE) > 0.000001) {            pre = cur;
    
     cur = (PRE/2 + x/(2*pre));        }        return int (cur);    }};
    

LeetCode69 Sqrt (x)