69. Sqrt(x)

Implement int sqrt(int x).

Compute and return the square root of x, where x is guaranteed to be a non-negative integer.

Since the return type is an integer, the decimal digits are truncated and only the integer part of the result is returned.

Example 1:

Input: 4
Output: 2

Example 2:

Input: 8
Output: 2
Explanation: The square root of 8 is 2.82842..., and since 
             the decimal part is truncated, 2 is returned.

 

my code:

class Solution {
public:
    int mySqrt(int x) {
        for (int i = 1; i <= x; i++) {
            if ((long)i*i > x)
                return i-1;
            else if ((long)i*i == x)
                return i;
        }   
        return 0;
    }
};

Runtime: 36 ms, faster than 8.63% of C++ online submissions for Sqrt(x).

 

2. Newton's Iterative Method in C++

首先，选择一个接近函数{\displaystyle f(x)}零点的{\displaystyle x_{0}}，计算相应的{\displaystyle f(x_{0})}和切线斜率{\displaystyle f'(x_{0})}（这里{\displaystyle f'}表示函数{\displaystyle f}的导数）。然后我们计算穿过点{\displaystyle (x_{0},f(x_{0}))}并且斜率为{\displaystyle f'(x_{0})}的直线和{\displaystyle x}轴的交点的{\displaystyle x}坐标，也就是求如下方程的解：

{\displaystyle 0=(x-x_{0})\cdot f'(x_{0})+f(x_{0})}

我们将新求得的点的{\displaystyle x}坐标命名为{\displaystyle x_{1}}，通常{\displaystyle x_{1}}会比{\displaystyle x_{0}}更接近方程{\displaystyle f(x)=0}的解。因此我们现在可以利用{\displaystyle x_{1}}开始下一轮迭代。迭代公式可化简为如下所示：

{\displaystyle x_{n+1}=x_{n}-{\frac {f(x_{n})}{f'(x_{n})}}}

已有证明牛顿迭代法的二次收敛^[1]必须满足以下条件：
{\displaystyle f'(x)\neq 0}; 对于所有{\displaystyle x\in I}，其中{\displaystyle I}为区间[α − r, α + r]，且{\displaystyle x_{0}}在区间其中{\displaystyle I}内，即 {\displaystyle r\geqslant \left|a-x_{0}\right|} 的; 
对于所有{\displaystyle x\in I}，{\displaystyle f''(x)}是连续的; 
{\displaystyle x_{0}}足够接近根 α。

 

class Solution {
public:
    int mySqrt(int x) {
        double ans = x;
        double delta = 0.0001;
        while (fabs(pow(ans, 2) - x) > delta) {
            ans = (ans + x / ans) / 2;
        }
        return (int)ans;
    }
};

Runtime: 24 ms, faster than 23.90% of C++ online submissions for Sqrt(x).

 

3. binary search:

class Solution {
public:
    int mySqrt(int x) {
        int low = 0,  high = x, mid;
        if(x<2) return x; // to avoid mid = 0
        while(low<high)
        {
            mid = (low + high)/2;
            if(x/mid >= mid) low = mid+1;
            else high = mid;
        }
        return high-1;
        
    }
};

　　

Runtime: 16 ms, faster than 53.68% of C++ online submissions for Sqrt(x).