問題描述

我需要編寫一個(gè)算法，它接受一個(gè)整數(shù)并返回所有可能的加法格式

I need to write an algorithm that takes an integer and returns all possible format of addition

例如

如果我進(jìn)入:6

它將返回以下字符串:

 0+6=6
 1+1+1+1+1+1=6
 1+1+1+1+2=6
 1+1+1+3=6
 1+1+4=6
 1+5=6
 2+1+1+1+1=6
 2+1+1+2=6
 2+1+3=6
 2+4=6
 3+1+1+1=6
 3+1+2=6
 3+3=6
 4+1+1=6
 4+2=6
 5+1=6
 6+0=6

這是我的嘗試:

import java.util.*;
public class Test
{
    public static void main(String[] args)
    {
        Scanner in = new Scanner(System.in);
        System.out.print("Enter an integer? ");
        int num = in.nextInt();
        System.out.println();
        calculate(num);
    }
    private static void calculate(int n)
    {
        int[] arInt = new int[n];
        for(int i = 0; i <= n; i++)
        {
            for(int j = 0; j <= n; j++)
            {
                arInt[j] = i;
            }
            // ...
        }
    }
}

推薦答案

我同意 Brad.完成此操作的最佳方法可能是通過遞歸.事實(shí)上，我昨晚正在研究與此相關(guān)的事情.我使用遞歸回溯算法解決了我的問題.查看維基百科頁面:回溯

I agree with Brad. The best way to complete this would probably be through recursion. In fact, I was working on something related to this last night. I solved my problem using a recursive backtracking algorithm. Check out the Wikipedia page: Backtracking

現(xiàn)在，我不保證沒有更好、更簡(jiǎn)單的方法來解決這個(gè)問題.但是，通過遞歸回溯，您將找到所有解決方案.

Now, I make no guarantees that there aren't better, less complex ways to solve this. However, with recursive backtracking you will find all the solutions.

但有一點(diǎn)需要注意，即 0.您可以將任意數(shù)量的零放入加法/減法中，結(jié)果會(huì)相同.

One thing to watch out for though, that 0. You can throw any amount of zeros into an addition/subtraction and it will come out the same.

這篇關(guān)于采用整數(shù)并返回所有可能的加法格式的算法的文章就介紹到這了，希望我們推薦的答案對(duì)大家有所幫助，也希望大家多多支持html5模板網(wǎng)！