【LeetCode & 剑指offer刷题】分治法题3：Sqrt(x)

【LeetCode & 剑指offer 刷题笔记】目录（持续更新中...）

 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.

---

C++

 

/*

本题其实与分治法无关

 

方法：变形的二分查找法，找最后一个不大于目标值（x/mid）的数（由于返回整数，故可以用此方法）

参见： 二分查找小结 - Grandyang - 博客园

*/

class Solution

{

public:

    int mySqrt(int x)

    {

        if (x <= 1) return x;

        int left = 0, right = x;

        while (left < right)

        {

            int mid = left + (right - left) / 2;

            if (mid <= (x/mid) )

                left = mid + 1;

            else

                right = mid;

        }

        return right - 1; //通过返回right-1,改造返回第一个大于目标值（x/mid）的数 -> 返回最后一个不大于目标值的数

    }

};

 

/*

方法二：牛顿迭代法

对于求方程f(x) = 0的根，可用迭代式 xk+1 = xk - f(xk)/f'(xk)进行求解

（特殊的简单迭代法xk+1 = p(xk),x = p(x)等价于f(x) = 0）

 

对此例x^2 = n，可推得xk+1 = (xk + n/xk)/2

*/

class Solution

{

public:

    int mySqrt(int n)

    {

        if (n <= 1) return n;

        

        double x = 1, x_pre = 0;

        while (abs(x - x_pre) > 1e-6) //精度具体依题目要求

        {

            x_pre = x;

            x = (x + n / x) / 2;

        }

        return int(x);

    }

};