Question

我正在尝试使用make一个方法，该方法接收两个字符串数组列表，并输出列表中任何一个列表中的值列表，但不能同时输出这两个值，不允许重复。

这是我迄今为止所做的，但我没有通过3次junit测试

public static ArrayList<String> getDifference(ArrayList<String> one, ArrayList<String>     two)
{
Set<String> oneSet = new LinkedHashSet<>(one);
ArrayList<String> finalone = new ArrayList<>(oneSet);
Set<String> twoSet = new LinkedHashSet<>(two);
ArrayList<String> finaltwo = new ArrayList<>(twoSet);
Collection<String> result = new ArrayList<String>(finaltwo);
result.removeAll(finalone);
ArrayList<String> list = new ArrayList<String>(result);
return list;
}

我失败的测试低于任何帮助，我可以做些什么来解决这个问题，我们将不胜感激，并提前感谢你。

    @Test
public void testGetDifferenceWithEmptyListSecond() {
    String[]          un   = { "luke", "leah", "han" };
    String[]          duex = { };
    ArrayList<String> one  = new ArrayList<String>( Arrays.asList( un ));
    ArrayList<String> two  = new ArrayList<String>( Arrays.asList( duex ));

    ArrayList<String> actual = Lab03Two.getDifference( one, two );

    assertEquals( "The number of elements is incorrect", 3, actual.size() );
    assertTrue  ( "The value \"luke\" was not found in the result", actual.contains( "luke" ));
    assertTrue  ( "The value \"leah\" was not found in the result", actual.contains( "leah" ));
    assertTrue  ( "The value \"han\" was not found in the result", actual.contains( "han" ));
}
    @Test
public void testGetDifferenceWithOverlapAndDuplicates() {
    String[]          un   = { "palpatine", "dooku", "vader", "sidius" };
    String[]          duex = { "padme", "vader", "sidius", "ackbar", "padme" };
    ArrayList<String> one  = new ArrayList<String>( Arrays.asList( un ));
    ArrayList<String> two  = new ArrayList<String>( Arrays.asList( duex ));

    ArrayList<String> actual = Lab03Two.getDifference( one, two );

    assertEquals( "The number of elements is incorrect", 4, actual.size() );
    assertTrue  ( "The value \"ackbar\" was not found in the result", actual.contains( "ackbar" ));
    assertTrue  ( "The value \"dooku\" was not found in the result", actual.contains( "dooku" ));
    assertTrue  ( "The value \"padme\" was not found in the result", actual.contains( "padme" ));
    assertTrue  ( "The value \"palpatine\" was not found in the result", actual.contains( "palpatine" ));
}
}

    @Test
public void testGetDifferenceWithNoOverlap() {
    String[]          un   = { "obi-wan", "jar-jar", "anakin" };
    String[]          duex = { "r2-d2", "c-3po" };
    ArrayList<String> one  = new ArrayList<String>( Arrays.asList( un ));
    ArrayList<String> two  = new ArrayList<String>( Arrays.asList( duex ));

    ArrayList<String> actual = Lab03Two.getDifference( one, two );

    assertEquals( "The number of elements is incorrect", 5, actual.size() );
    assertTrue  ( "The value \"obi-wan\" was not found in the result", actual.contains( "obi-wan" ));
    assertTrue  ( "The value \"jar-jar\" was not found in the result", actual.contains( "jar-jar" ));
    assertTrue  ( "The value \"anakin\" was not found in the result", actual.contains( "anakin" ));
    assertTrue  ( "The value \"r2-d2\" was not found in the result", actual.contains( "r2-d2" ));
    assertTrue  ( "The value \"c-3po\" was not found in the result", actual.contains( "c-3po" ));
}

Answer 1

你的逻辑有点偏。

设置联盟：{A，B}∪{B，C} = {A，B，C} [所有没有复制的元素]
设置交叉点：{A，B}∩{B，C} = {B} [常用元素] 设置差异：{A，B} / {B，C} = {A} [通知，无C元素]

你想要设置联盟 - 设置交集 :(对称差异）信用这里归功于Phil

{“palpatine”，“dooku”，“vader”，“sidius”}
{“padme”，“vader”，“sidius”，“ackbar”，“padme”}

（1∪2/1∩2）

{palpatine dooku vader sidius padme ackbar} - {padme vader} = {palpatine dooku sidius ackbar}

image example

replaceAll应用set 差异 :( oneSet difference twoSet）

旅程来自 removeAll 方法。它会从一个中删除两个所有元素，但不会添加一个中不的所有元素>到两个。

以下代码通过执行 2 set difference和1 union 来修复测试，以实现对称差异/独占或。

独家或（xor），对称差异

public static List<String> getXOR(List<String> oneArray, List<String> twoArray) {

Set<String> oneSet = new HashSet<>(oneArray);
Set<String> twoSet = new HashSet<>(twoArray);

oneSet.removeAll(twoArray);// 1. oneSet / twoArray    ,  oneSet !AND twoArray
twoSet.removeAll(oneArray);// 2. twoSet / oneArray    ,  twoSet !AND oneArray
oneSet.addAll(twoSet);     // 3. oneSet U twoSet      ,  oneSet OR   twoSet

return new ArrayList<String>(oneSet);
}

Answer 2

public static ArrayList<String> getDifference(ArrayList<String> one, ArrayList<String> two)
{
    ArrayList<String> list = new ArrayList<String>();

    //iterate over all elements of one
    //if two does not contain it, it's a difference -> add it
    for (String i : one) {
        if (!two.contains(i)) {
            set.add(i);
        }
    }

    //same with two
    for (String i : two) {
        if (!one.contains(i)) {
            set.add(i);
        }
    }

    return list;
}

Answer 3

嗯，这里最简单的解决方案是遍历两个列表，检查单词是否存在于另一个列表中，如果不存在，则将其添加到Set中。

Set<String> solution = new LinkedHashSet<>();
int n1 = one.size();
int n2 = two.size();
int i=0;
int j=0;
while (i<n1) {
    while (j<n2) {
       if (!one.get(i).equals(two.get(j))) 
            solution.add(one.get(i));
       j++;
    }

    if (i>=n2) {
      solution.add(one.get(i));
    }
    i++;
}

while (j<n2) {
  solution.add(two.get(j));
  j++;
}

getDifference（ArrayList <string> one，ArrayList <string> two）方法</string> </string>

3 个答案: