如何在不使用Java的reduce方法的情况下获得相同的结果?

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英文:

How to get same result without using Java's reduce method?

问题

我有一个方法,应该能够计算多个数字的最小公倍数(LCM)。它使用了 Java 的 reduce() 方法,因此数字 1、2、3 会得到正确的最小公倍数 6:

int lcmAnswer = Arrays.stream(numbers).reduce(1, (a, b) -> {
            int total = lcmm(a, b);
            return total;
    }
);
System.out.println(lcmAnswer); // LCM(1, 2, 3) = 6

然而,如果不使用 Java 的 reduce() 方法,那么数字 1、2、3 就无法得到最小公倍数 6。而是得到了错误的最小公倍数 LCM(1, 2, 3) = 8:

int[] numbers = {1, 2, 3};
System.out.println(lcmm(1, 2, 3)); // LCM(1, 2, 3) = 8,这是错误的

private static int lcmm(int... numbers) {
    int sum = 0;
    for (int i = 0; i < numbers.length - 1; i++) {
        int curr = numbers[i];
        int next = numbers[i + 1];
        sum += lcm(curr, next);
    }

    return sum;
}

private static int lcm(int p, int q) {
    // 返回最小公倍数
    return p * q / gcd(p, q);
}

private static int gcd(int p, int q) {
    // 使用欧几里得算法返回最大公约数
    int temp;
    while (q != 0) {
        temp = q;
        q = p % q;
        p = temp;
    }
    return p;
}

有人知道我做错了什么吗?

英文:

I have method which should give the LCM of multiple numbers. It works with Java's reduce() method so the numbers 1,2,3 gives the LCM of 6, which is correct:

int lcmAnswer = Arrays.stream(numbers).reduce(1, (a, b) -&gt; {
            int total =  lcmm(a, b);
            return total;
    }
);
System.out.println(lcmAnswer); // LCM(1, 2, 3) = 6

However if I don't use Java's reduce() method then the numbers 1,2,3 don't give me the LCM of 6. It gives me LCM(1, 2, 3) = 8, which is wrong:

int[] numbers = {1, 2, 3};
System.out.println(lcmm(1,2,3)); // LCM(1, 2, 3) = 8, which is wrong


private static int lcmm(int... numbers) {
    int  sum = 0;
    for (int i = 0; i&lt;numbers.length -1; i++) {
        int curr = numbers[i];
        int next = numbers [i+1];
        sum += lcm(curr, next);
    }

    return sum;
}

private static int lcm(int p, int q) {
    // Return lowest common multiple.
    return p * q / gcd(p, q);
}

private static int gcd(int p, int q) {
    //Return greatest common divisor using Euclid&#39;s Algorithm.
    int temp;
    while (q != 0) {
        temp = q;
        q = p % q;
        p = temp;
    }
    return p;
}

Does someone has any idea what I'm doing wrong?

答案1

得分: 1

假设我们有四个数字 a,b,c,d。要计算 a,b,c,d 的最小公倍数(LCM),需要按照以下步骤进行:

设最终的 LCM 为 RES

  • 计算前两个数字 ab 的 LCM。将值赋给 RES。因此,RES = LCM(a,b)
  • 计算 RESc 的 LCM。更新 RES 的值。因此,RES = LCM(RES,c)
  • 计算 RESd 的 LCM。更新 RES 的值。因此,RES = LCM(RES,d)

最终的 RES 值将包含 a,b,c,d 的 LCM。

我们可以按照这个算法计算多个数字的 LCM。

以下是实现代码:

import java.util.*;
import java.lang.*;
import java.io.*;

class LCMMultiple{
    public static void main (String[] args) throws java.lang.Exception{
        int[] numbers = {1, 2, 3};
        System.out.println(getLcmMultiple(numbers));

    }

    private static int getLcmMultiple(int... numbers) {
        int lcm = 0;
        for (int i = 0; i<numbers.length -1; i++) {
            int curr = numbers[i];
            int next = numbers [i+1];
            if(lcm != 0){
                lcm = getLcm(lcm, getLcm(curr, next));
            }
            else{
                lcm = getLcm(curr, next);
            }
        }
        return lcm;
    }

    private static int getLcm(int p, int q) {
        // 返回最小公倍数。
        return p * q / getGcd(p, q);
    }

    private static int getGcd(int p, int q) {
        // 使用欧几里德算法返回最大公约数。
        int temp;
        while (q != 0) {
            temp = q;
            q = p % q;
            p = temp;
        }
        return p;
    }	 
}

输出结果:

6
英文:

Suppose we have four numbers a,b,c,d. To calculate LCM of a,b,c,d, we need to follow these steps:

Let the final LCM is RES.

  • Calculate LCM of first two numbers a and b. Assign the value to RES. So, RES = LCM(a,b).
  • Calculate LCM of RES and c. Update the value of RES. So, RES = LCM(RES,c).
  • Calculate LCM of RES and d. Update the value of RES. So, RES = LCM(RES,d).

The final value of RES will contain the LCM of a,b,c,d.

We can follow this algorithm to calculate LCM of multiple numbers.

Here is the implementation:

import java.util.*;
import java.lang.*;
import java.io.*;
 
class LCMMultiple{
	public static void main (String[] args) throws java.lang.Exception{
		int[] numbers = {1, 2, 3};
		System.out.println(getLcmMultiple(numbers));
 
	}
 
	private static int getLcmMultiple(int... numbers) {
	    int lcm = 0;
	    for (int i = 0; i&lt;numbers.length -1; i++) {
	        int curr = numbers[i];
	        int next = numbers [i+1];
	        if(lcm != 0){
		        lcm = getLcm(lcm, getLcm(curr, next));
	        }
	        else{
	        	lcm = getLcm(curr, next);
	        }
	    }
	    return lcm;
	}
 
	private static int getLcm(int p, int q) {
	    // Return lowest common multiple.
	    return p * q / getGcd(p, q);
	}
 
	private static int getGcd(int p, int q) {
	    //Return greatest common divisor using Euclid&#39;s Algorithm.
	    int temp;
	    while (q != 0) {
	        temp = q;
	        q = p % q;
	        p = temp;
	    }
	    return p;
	}	 
}

Output:

6

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  • 本文由 发表于 2020年8月18日 04:54:14
  • 转载请务必保留本文链接:https://go.coder-hub.com/63458507.html
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