使用二分搜索查找给定数的立方根的Java程序

给定一个非负数，使用二进制搜索方法找到一个数的立方根。

例子 ：

Input: x = 27
Output: 3
Explanation:  The cube root of 16 is 4.

Input: x = 120
Output: 4
Explanation:  The cube root of 120 lies in between
4 and 5 so floor of the cube root is 4.

天真的方法：

• 检查每个元素的立方体直到 n 并存储答案直到立方体小于或等于 n

Java

// Java Program to Find the cube root
// of given number using Naive approach
import java.io.*;
class GFG {
    static int cuberoot(int n)
    {
        int ans = 0;
  
        for (int i = 1; i <= n; ++i) {
            // checking every number cube
            if (i * i * i <= n) {
                ans = i;
            }
        }
        return ans;
    }
    public static void main(String[] args)
    {
        // Number
        int number = 27;
        // Checking number
        int cuberoot = cuberoot(number);
        System.out.println(cuberoot);
    }
}

输出

3

复杂：

SpaceComplexity: O(1)
TimeComplexity: O(n)

高效方法（二分搜索）：

二分搜索使用分而治之的方法，使得复杂度为 O(log n)。

算法：

• 初始化 left=0 和 right =n
• 计算mid=left+(right-left)/2
• 如果 mid*mid*mid 等于数字返回 mid
• 如果 mid*mid*mid 小于数字，则将 mid 存储在 ans 中并增加 left=mid+1
• 如果 mid*mid*mid 大于数字并减少 right=mid-1
• 返回答案

执行：

Java

// Java Program to Find the cube root
// of given number using Binary Search
import java.io.*;
import java.util.*;
class GFG {
    // Function to find cuberoot
    static int cuberoot(int number)
    {
        // Lower bound
        int left = 1;
        // Upper bound
        int right = number;
  
        int ans = 0;
        while (left <= right) {
            // Finding the mid value
  
            int mid = left + (right - left) / 2;
            // Checking the mid value
            if (mid * mid * mid == number) {
                return mid;
            }
  
            // Shift the lower bound
            if (mid * mid * mid < number) {
                left = mid + 1;
                ans = mid;
            }
            // Shift the upper bound
            else {
                right = mid - 1;
            }
        }
        // Return the ans
        return ans;
    }
    public static void main(String[] args)
    {
        int number = 215;
        System.out.println(cuberoot(number));
    }
}

输出

5

复杂：

SpaceComplexity: O(1)
TimeComplexity: O(log n)