ArrayList中字符串的置换(使用递归)

Question

我已经试着弄清楚这个问题有一段时间了，只是不太明白我是如何创造预期的结果的。我已经成功地获得了唯一排列的输入和最大数量，现在我所需要的只是找出如何获得期望的字符串作为输出。我认为有更好的方法去做这件事。

投入: Julia Lucas Mia -1

预期产出：

茱莉亚，卢卡斯，米娅

朱莉娅，米娅，卢卡斯

卢卡斯，朱莉娅，米娅

卢卡斯，米娅，朱莉娅

米娅，朱莉娅，卢卡斯

米娅，卢卡斯，茱莉亚

import java.util.Scanner;
import java.util.ArrayList;

public class PhotoLineups {

    // TODO: Write method to create and output all permutations of the list of names.
        public static void printAllPermutations(ArrayList<String> permList, ArrayList<String> nameList){
        int size = nameList.size();
        int index = 0;

        for(int i = size; i > 1; i--)
            size = size * (i - 1);
        System.out.println("Size: " +size); //Temp

        if(index != size){
            for(int i = 0; i < nameList.size(); i++){
                
                index++;
            }
        }
    }

    public static void main(String[] args){
        Scanner scnr = new Scanner(System.in);
        ArrayList<String> nameList = new ArrayList<String>();
        ArrayList<String> permList = new ArrayList<String>();
        String name;
        int i = 0;
        
        // TODO: Read in a list of names; stop when -1 is read. Then call recursive method.
        while(i != -1){
            name = scnr.next();
            nameList.add(name);
            if(name.equals("-1"))
                i = -1;
        }
        nameList.remove(nameList.size() - 1);
        System.out.println("nameList: " + nameList); //Temp

        //Recursive
        printAllPermutations(permList, nameList);
    }
}

Answer 1

在这种情况下，递归意味着将问题一分为二。因此，算法将在循环中选择列表的名称，并将其与其他列表的排列组合起来。如果其余的是空的，你就得到了一个排列:你必须复制列表，因为它们是通过引用传递的.

public static void printAllPermutations(ArrayList<String> permList, ArrayList<String> nameList){

    if (nameList.isEmpty()) {
        System.out.println(String.join(", ", permList));
        return;
    }

    for(int i = 0; i < nameList.size(); i++) {
        ArrayList<String> localPermList = new ArrayList<>(permList);
        ArrayList<String> localNameList = new ArrayList<>(nameList);
        localPermList.add(nameList.get(i));
        localNameList.remove(nameList.get(i));
        printAllPermutations(localPermList, localNameList);

    }

}

编辑

复制yout代码，最好将循环编写为

    for(String name : nameList) {
        ArrayList<String> localPermList = new ArrayList<>(permList);
        ArrayList<String> localNameList = new ArrayList<>(nameList);
        localPermList.add(name);
        localNameList.remove(name);
        printAllPermutations(localPermList, localNameList);

    }