Sqrt(x) Solutions in C++

// Source : https://oj.leetcode.com/problems/sqrtx/
// Author : Hao Chen
// Date   : 2014-08-26



#include <stdlib.h>
#include <iostream>
using namespace std;


int sqrt(int x) {

    if (x <=0 ) return 0;
    
    //the sqrt is not greater than x/2+1
    int e = x/2+1;
    int s = 0;
    // binary search
    while ( s <= e ) {
        int mid = s + (e-s)/2;
        long long sq = (long long)mid*(long long)mid;
        if (sq == x ) return mid;
        if (sq < x) {
            s = mid + 1;
        }else {
            e = mid - 1;
        }
    }
    return e; 
        
}

// http://en.wikipedia.org/wiki/Newton%27s_method
int sqrt_nt(int x) {
    if (x == 0) return 0;
    double last = 0;
    double res = 1;
    while (res != last)
    {
        last = res;
        res = (res + x / res) / 2;
    }
    return int(res);
}


int main(int argc, char**argv)
{
    int n = 2;
    if( argc > 1 ){
        n = atoi(argv[1]);
    }
    cout << "sqrt(" << n << ") = " << sqrt(n) << endl;
    return 0;
}

class Solution {
public:
    int sqrt(int x) {
        int l = 1, r = x, ret = 0;
        while (l <= r) {
            int m = (l + r) >> 1;
            if (m <= x / m) { l = m+1; ret = m; }
            else r = m-1;
        }
        return ret;
    }
};

/// Source : https://leetcode.com/problems/sqrtx/description/
/// Author : liuyubobobo
/// Time   : 2018-06-18

#include <iostream>

using namespace std;


/// Binary Search
/// Time Complexity: O(log(MAX_INT))
/// Space Complexity: O(1)
class Solution {
public:
    int mySqrt(int x) {

        int l = 0, r = x;
        while(l < r){
            long long mid = l + ((long long)r - l + 1) / 2;
            if(mid * mid <= (long long)x)
                l = mid;
            else
                r = mid - 1;
        }
        return l;
    }
};


int main() {

    cout << Solution().mySqrt(4) << endl;
    cout << Solution().mySqrt(8) << endl;

    return 0;
}

/// Source : https://leetcode.com/problems/sqrtx/description/
/// Author : liuyubobobo
/// Time   : 2019-04-03

#include <iostream>

using namespace std;


/// Binary Search
/// Using double first
///
/// Time Complexity: O(log(MAX_INT) * precision)
/// Space Complexity: O(1)
class Solution {

private:
    double e = 1e-6;

public:
    int mySqrt(int x) {

        double l = 0.0, r = INT_MAX;
        while(r - l >= e){
            double mid = (l + r) / 2;
            if(mid * mid <= x)
                l = mid;
            else
                r = mid;
        }
        return (int)r;
    }
};


int main() {

    cout << Solution().mySqrt(4) << endl;
    cout << Solution().mySqrt(8) << endl;

    return 0;
}