sqrt函数实现

这个题只想到二分法写，

private static int mysqrt(int x) {
        if(x<2) return x;
        /*
         利用二分查找做，一般start = 2;end =  x/2 + 1,因为end*end = (x^2/4+x+1) 
         */
        int start = 2;
        int end = x/2 + 1;
        int mid = (end - start) / 2 + start;
        while(start < end) {
            if(mid  > x / mid) {
                end = mid;
            }else if(mid == x / mid) {
                return mid;
            }else {
                start = mid + 1;
            }
            mid = (end - start) / 2 + start;
        }
        return mid - 1;

另外看了他人实现的sqrt方法，感觉挺好的，下面附上博客地址：

https://blog.csdn.net/gqtcgq/article/details/48392255