说明

算法：Sqrt(x)
LeetCode地址：https://leetcode.com/problems/sqrtx/

题目：
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.

解题思路1

求平方根，思路为从缩小两端范围[left, right], left为从1开始，right为从入参x开始。条件如下：

x < 1, 直接返回0
取中值 mid = left + right >> 1, 用位移的方式来获取除以2更快速
如果mid > x / mid, 则缩小right = mid
如果mid <= x / mid, 则(mid + 1) * (mid + 1) > x, 则说明mid为最终结果；否则，left = mid + 1;

代码实现1

public class Sqrt {

    public static int mySqrtWithMid(int x) {
        if(x < 1) {
            return 0;
        }
        int left = 0;
        int right = x;
        while(true) {
            int mid = left + ((right - left) >> 1);
            if (mid < 1) {
                return 1;
            }
            if (mid > x / mid) {
                right = mid;
            } else {
                if (mid + 1 > x/(mid + 1)) {
                    return mid;
                }

                left = mid + 1;
            }
        }
    }

    public static void main(String[] args) {
        int input1 = 4;
        System.out.println("input: " +input1+ " mySqrtWithMid: " + mySqrtWithMid(input1));

        int input2 = 8;
        System.out.println("input: " +input2+ " mySqrtWithMid: " + mySqrtWithMid(input2));
    }
}

运行结果1

input: 4 mySqrtWithMid: 2
input: 8 mySqrtWithMid: 2

代码执行效率1

Runtime: 13 ms, faster than 99.97% of Java online submissions for Sqrt(x).
Memory Usage: 38.1 MB, less than 7.02% of Java online submissions for Sqrt(x).

解题思路2

第二种思路需要证明, 喜欢开脑洞的可以看看参考链接如何证明，参考牛顿迭代法求平方根，然后再参考维基百科的 Integer square root 即可。
在这里插入图片描述

代码实现2

public class Sqrt {

    public static int mySqrt(int x) {
        if (x < 1) {
            return 0;
        }

        long y = x;
        while (y * y > x) {
            y = (long)(0.5 * (y + x / y));
        }

        return (int)y;
    }

    public static void main(String[] args) {
        int input1 = 4;
        System.out.println("input: " +input1+ " mySqrt: " + mySqrt(input1));

        int input2 = 8;
        System.out.println("input: " +input2+ " mySqrt: " + mySqrt(input2));
    }
}

运行结果2

input: 4 mySqrt: 2
input: 8 mySqrt: 2

代码执行效率2

Runtime: 14 ms, faster than 94.36% of Java online submissions for Sqrt(x).
Memory Usage: 38.1 MB, less than 5.29% of Java online submissions for Sqrt(x).
Next challenges:

总结

二分法应用。

代码下载：
https://github.com/zgpeace/awesome-java-leetcode/blob/master/code/LeetCode/src/Sqrt.java

算法:Sqrt(x) (x 的平方根)

说明