通过递归

Question

我已经在本网站上搜索了许多类似类似基本概念的问题，但是经过许多小时试图自己解决这个问题并审核我是仍然输了。如果还有另一个问题可以回答这个问题，我将非常乐意为您介绍一下。

最终我想创建一个递归方法，这样它需要两个列表并返回一组字符串列表：

//Example of such a function definition
private static Set<List<String>> myRecursiveMethod(List<String> listOne, 
List<String> listTwo) {
}

当我说＆＃34;字符串列表＆＃34;我的意思是具体以下内容： （注意：＆＃34; AD＆＃34; ==＆＃34; DA＆＃34;）

// if the two following lists are INPUTTED into myRecursiveMethod();
// listOne = ["A","B"]
// listTwo = ["C","D"]
// the following set is OUTPUTTED: [["AC","BD"],["AD","BC"]] 

这样如果listOne和listTwo中都有三个元素，那么集合中就会有SIX元素。即：

// listOne = ["A","B","C"]
// listTwo = ["D","E","F"]
// OUTPUTTED: [["AD","BE","CF"],["AD","BF","CE"],["BD","AE","CF"],
// ["BD","AF","CE"],["CD","AE","BF"],["CD","AF","BE"]] 

我尝试使用双增强FOR循环来编写它，这样我就能理解逻辑。我的FOR循环方法非常糟糕，只适用于list.size（）== 2的HARD-CODED限制。

// Create Lists and append elements 
List<String> listOne = new ArrayList<String>();
listOne.add("A");
listOne.add("B");

List<String> listTwo = new ArrayList<String>();
listTwo.add("C");
listTwo.add("D");

// List One = ["A","B"]
// List Two = ["C","D"]

// Create new List
List<List<String>> newList = new ArrayList<List<String>>();
Integer counter = 0;

    for (String s : listOne) {
        counter++;
        for (String p : listTwo) {

            // A HARD-CODED bad implementation of this method
            if (counter < 3) {
                List<String> newListTwo = new ArrayList<String>();
                newListTwo.add(s.concat(p));
                newList.add(newListTwo);

            } else if (!(counter % 2 == 0)) {
                newList.get(1).add(s.concat(p));

            } else {
                newList.get(0).add(s.concat(p));
            }
        }
    }
    System.out.println(newList); // = [["AC","BD"],["AD","BC"]] 

另外你可以注意到我定义了List＆lt; List＆lt; String＆gt;＆gt; 而不是设置＆lt; List＆lt; String＆gt;＆gt;。这是由于我的编码错误，它依赖于list.get（）方法。

所以我目前的递归方法如下：

private static Set<List<String>> myRecursiveMethod(List<String> listOne, 
List<String> listTwo) 
{
    //Base Case:
    if (listOne.isEmpty){
        return new HashSet<List<String>>;
    }
    //Recursive Case:
    else {
        String listOneFirst = listOne.get(0);
        String listTwoFirst = listTwo.get(0);

        List<String> sampleList = new ArrayList<String>();
        sampleList.add(listOneFirst+listTwoFirst);

        Set<List<String>> newSet = new HashSet<List<String>>(myRecursiveMethod())
        newSet.add(sampleList);
        return newSet;
    }
}

此方法目前只有这样：

INPUT：

• 列表一= [＆＃34; A＆＃34;，＆＃34; B＆＃34;]
• 清单二= [＆＃34; C＆＃34;，＆＃34; D＆＃34;]

输出：

• [[＆＃34; AC＆＃34;] [＆＃34; BD＆＃34;]]

期望的输出：

• [[＆＃34; AC＆＃34;＆＃34; BD＆＃34;]，[＆＃34; AD＆＃34;＆＃34; BC＆＃34;]]

修改

在审核了我的W.I.P代码后的答案：

private static Set<List<String>> myRecursiveMethod(List<String> listOne,
        List<String> listTwo) {

    //Backup Case (user enters an empty list)
    if (listOne.isEmpty()){
        return new HashSet<List<String>>();
    }

    // Base Case:
    if (listOne.size() == 1) {
        List<String> mergedStrings = new ArrayList<>();
        for (String s : listTwo) {
            mergedStrings.add(listOne.get(0).concat(s));
        }
        Set<List<String>> builtHashSet = new HashSet<List<String>();
        builtHashSet.add(mergedStrings);
        return builtHashSet;
    }
    // Recursive Case:
    else {

        // Ensure original list values arn't changed.
        List<String> newListOne = new ArrayList<String>(listOne);
        List<String> newListTwo = new ArrayList<String>(listTwo);

        //first two elements...I don't think this is correct
        String listOneFirst = newListOne.get(0);
        String listTwoFirst = newListTwo.get(0);
        List<String> sampleList = new ArrayList<String>();
        sampleList.add(listOneFirst + listTwoFirst);

        //used for making recursive case smaller
        newListOne.remove(0);

        // Calls recursion
        Set<List<String>> newSet = new HashSet<List<String>>(
                myRecursiveMethod(newListOne, newListTwo));
        newSet.add(sampleList);

        return newSet;
    }
}