是否有一种算法可以删除与另一个元素相同的元素？

Question

我想删除一些与其他输出值相同的输出！因为Test1=Test2=Test5=Test6，所以我希望它在控制台上只显示Test1！Test3=Test4=Test7=Test8，所以我希望它只显示Test3。。。。。。。。

以下代码是我所做的：

    boolean flg1 = true;
    List<List<String>> tmp1 = new ArrayList <>();
    List<List<String>> tmp2 = new ArrayList <>();
    System.out.println();
    System.out.println("Size of the subset: " + subset.size());
    for(int i=0; i< subset.size();i++){
        tmp1=subset.get(i);

        // System.out.println("TMP1:"+ tmp1);

        for(int j=0; j<subset.size();j++){
        tmp2=subset.get(j);

        // System.out.println("TMP2:"+ tmp2.get(1));

            if(i != j && !compareOuterlist(tmp1,tmp2)){
                flg1=true;
                continue;
            }
        }
        if(flg1){
            count++;
            System.out.println("TEST " + (i+1) + ":"+ tmp1);
        }
    }

    System.out.println();
    System.out.println("Total partition:"+ count);
}
public static boolean compareOuterlist(List<List<String>> p1, List<List<String>> p2){
    if(p1.size() != p2.size()){
        return false;
    }
    for(int i = 0;i < p1.size();i++){
        if(!(compareInnerlist1(p1.get(i),p2.get(i)))){
            return false;
        }
    }
    return true;
}

public static boolean compareInnerlist1(List<String> p1, List<String> p2){
    for(int i=0;i<p1.size();i++){
        if(!(p1.equals(p2))){
            return false;
        }
    }
    return true;
}

以下是我的输出：

TEST 1:[[P11, X11, X22], [P11, X11, X22]]
TEST 2:[[P11, X11, X22], [P11, X11, X22]]
TEST 3:[[X11, X22], [P11, X22], [P11, X11]]
TEST 4:[[X11, X22], [P11, X22], [P11, X11]]
TEST 5:[[P11, X11, X22], [P11, X11, X22]]
TEST 6:[[P11, X11, X22], [P11, X11, X22]]
TEST 7:[[X11, X22], [P11, X22], [P11, X11]]
TEST 8:[[X11, X22], [P11, X22], [P11, X11]]
TEST 9:[[P11, X22], [X11, X22], [P11, X11]]
TEST 10:[[P11, X22], [X11, X22], [P11, X11]]
TEST 11:[[P11, X22], [X11, X22], [P11, X11]]
TEST 12:[[P11, X22], [X11, X22], [P11, X11]]

Answer 1

只要您想打印测试，您就可以创建字符串an的ArrayList，将结果添加到该列表中，但在打印和递增计数之前，您必须确保要打印的内容不存在于列表中：

if(flg1){
        if !(Your_List.conatains("TEST " + (i+1) + ":"+ tmp1)) {
             System.out.println("TEST " + (i+1) + ":"+ tmp1);
             count++;
        }
        Your_List.add("TEST " + (i+1) + ":"+ tmp1);
    }

Answer 2

使用Java8 Streams，您可以使用准确度（）来完成：

List<List<List<String>>> result = subset.stream().distinct().collect(Collectors.toList());

如果您想在打印时循环执行，可以使用Set完成：

Set<List<List<String>>> distinct = new HashSet<>();
for (int i = 0; i < subset.size(); i++) {
    List<List<String>> tmp1 = subset.get(i);
    if (distinct.add(tmp1))
        System.out.println("TEST " + (i+1) + ":"+ tmp1);
}
System.out.println();
System.out.println("Total partition: " + distinct.size());

输出

TEST 1:[[P11, X11, X22], [P11, X11, X22]]
TEST 3:[[X11, X22], [P11, X22], [P11, X11]]
TEST 9:[[P11, X22], [X11, X22], [P11, X11]]

Total partition: 3