[leetcode] 69 Sqrt(x)_myjayus_

问题描述：

Implement int sqrt(int x).

Compute and return the square root of x.

基本思路：

采用二分查找法； 犹豫输入x是int型 ，int最大范围是2147483647 ，所以返回结果的范围0-46340. 注意要根据确定的区间去查找正确的返回值，如果大于46340的数的平方将超出int的表示范围。

代码：

int sqrt(int x) {
        if(x==0||x==1)
            return x;
        if(x >=2147395600)
            return 46340;
        int high = 46339*2-1;
        int low  = 1;
        
        while(low <= high){
            int mid = (high+low)/2;
            int hpow = (mid+1)*(mid+1);
            int lpow = mid*mid;
            if(x>=lpow && x<hpow)
                return mid;
            else if(x<lpow)
                high = mid-1;
            else 
                low = mid+1;
        }
        
    }

a better method

Since sqrt(x) is composed of binary bits, I calculate sqrt(x) by deciding every bit from the most significant to least significant. 

 public int sqrt(int x) {
    if(x==0)
        return 0;
    int h=0;
    while((long)(1<<h)*(long)(1<<h)<=x) // firstly, find the most significant bit
        h++;
    h--;
    int b=h-1;
    int res=(1<<h);
    while(b>=0){  // find the remaining bits
        if((long)(res | (1<<b))*(long)(res |(1<<b))<=x)
            res|=(1<<b);
        b--;
    }
    return res;
}