leetcode Question 102: Sqrt(x)

Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x.

Analysis:
According to Newton's Method(http://en.wikipedia.org/wiki/Newton's_method), we can use

to get the sqrt(x).

Code:

class Solution {
public:
    int sqrt(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (x==0) {return 0;}
        if (x==1) {return 1;}
      
        double x0 = 1;
        double x1;
        while (true){
            x1 = (x0+ x/x0)/2;
            if (abs(x1-x0)<1){return x1;}
            x0=x1;
        }
        
    }
};

Another solution: the binary search approach is a more general way of solving this problem. One thing you need to consider is the length of the input, since taking the mid of a big value and computing its square may overflow the int type.
We can use "long long" , which have a max value 2^63-1.
The max of an int is 2^15-1
The max of a long is 2^31-1

class Solution {
public:
    int sqrt(int x) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        long long high = x;
        long long low = 0;
        if (x<=0) {return 0;}
        if (x==1) {return 1;}
        while (high-low >1){
            long long mid = low + (high-low)/2;
            if (mid*mid<=x){low = mid;}
            else {high = mid;}
        }
        return low;
    }
};