Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

思路：binary search ，柯西不等式 √ab<=(a+b)/2;

注意：overflow问题。新技巧：如果是乘法引起的overflow，可以试试改成除法。！！so clever！

另一种方法是用 位移运算。十分巧妙。原理在《位移运算》一章讲解 http://www.cnblogs.com/renrenbinbin/p/4337552.html

class Solution {
public:
    /*overflow must be considered*/
    int sqrt(int x) {
        int a=1;
        int b=x;
        int mid=0;
        while(a<b){
            mid=(a+b)/2;
            if(mid>x/mid) b=mid-1;
            else a=mid+1;
        }
        
        if(a*a>x) return a-1;
        else return b;
        
    }
};

方法2

class Solution {
public:
    /*overflow must be considered*/
    int sqrt(int x) {
        if(x==0) return 0;
        int a=1;
        int n=0;
        while((a<<n)<=(x/(a<<n))){
            n++;    
        }
        n--;
        int rel = a<<n;
        int b=n-1;
        while(b>=0){
            int m=1<<b;
            if((rel|m)<=x/(rel|m)){
                rel|=m;
            }
            b--;
        }
        return rel;
    }
};