[LintCode] 586 Sqrt(x) II 解题报告

Description
Implement double sqrt(double x) and x >= 0.

Compute and return the square root of x.

Notice
You do not care about the accuracy of the result, we will help you to output results.


Example
Given n = 2 return 1.41421356

5/2/2017

算法班

注意第10-12行，当x小于1时，mid只能无限接近x，而不会等于或者超过x。而在这个范围内sqrt(x) > x，所以需要把end设为1

（或者判断范围改变循环里面？？？）

public class Solution {
    /**
     * @param x a double
     * @return the square root of x
     */
    public double sqrt(double x) {
        // Write your code here
        double start = 0;
        double end = x;
        if (end < 1) {
            end = 1;
        }

        while (end - start > 1e-12) {
            double mid = (start + end) / 2;
            if (mid * mid < x) {
                start = mid;
            } else {
                end = mid;
            }
        }
        return start;
    }
}