删除字符串数组中具有相同字符的字符串

Question

我现在面临一个问题。在我的一个程序中，我需要从数组中删除具有相同字符的字符串。例如。假设，

我有 3 个类似的数组，

String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"};
String[] name2 = {"alan", "may", "jeremy", "helen", "alexi"};
String[] name3 = {"adel", "aron", "amy", "james", "yam"};

如您所见，name1 数组中有一个字符串 amy。另外，我在接下来的两个数组中有may、amy 和yam 之类的字符串。我需要的是，我需要一个不包含这些重复字符串的最终数组。我只需要出现一次：我需要删除最终数组中名称的所有排列。那就是最终的数组应该是：

String[] finalArray={"amy", "jose", "alice", "patrick","alan", "jeremy", "helen", "alexi","adel", "aron", "james"}

（上面的数组删除了 yam，may，并且只包括 amy）。

到目前为止，我使用HashSet 所做的尝试如下

String[] name1 = {"Amy", "Jose", "Jeremy", "Alice", "Patrick"};
String[] name2 = {"Alan", "mAy", "Jeremy", "Helen", "Alexi"};
String[] name3 = {"Adel", "Aaron", "Amy", "James", "Alice"};
Set<String> letter = new HashSet<String>();
for (int i = 0; i < name1.length; i++) {
    letter.add(name1[i]);
}
for (int j = 0; j < name2.length; j++) {
    letter.add(name2[j]);
}
for (int k = 0; k < name3.length; k++) {
    letter.add(name3[k]);
}
System.out.println(letter.size() + " letters must be sent to: " + letter);

但是，这段代码的问题在于，它只是删除了同一字符串的多次出现。还有其他选择吗？非常感谢任何帮助。

Answer 1

您可以对字符串 (str.toCharArray ()) 的字符数组进行排序，并从排序后的数组中创建一个新字符串，以获得字符串的“规范”表示。

然后您可以将这些字符串添加到Set，并检查每个字符串的规范表示是否已经在集合中。

Set<String> letter = new HashSet<String>();
for (int i = 0; i < name1.length; i++) {
    char[] chars = name1[i].toCharArray();
    Arrays.sort(chars);
    letter.add(new String(chars));
}
for (int j = 0; j < name2.length; j++) {
    char[] chars = name2[j].toCharArray();
    Arrays.sort(chars);
    letter.add(new String(chars));
}
for (int k = 0; k < name3.length; k++) {
    char[] chars = name3[k].toCharArray();
    Arrays.sort(chars);
    letter.add(new String(chars));
}

编辑：我将Set<char[]> 更改为Set<String>，因为数组不会覆盖hashCode 和equals，所以HashSet<char[]> 不起作用。

Answer 2

TreeSet 允许我们提供一个比较器。看看这是否有帮助。为了保持计数，请使用TreeMap。

package empty;

import java.util.Arrays;
import java.util.Comparator;
import java.util.Set;
import java.util.TreeMap;
import java.util.TreeSet;

public class RemoveDuplicateStrings {

    public static void main(String[] args) {
        String[] name1 = { "amy", "jose", "jeremy", "alice", "patrick" };
        String[] name2 = { "alan", "may", "jeremy", "helen", "alexi" };
        String[] name3 = { "adel", "aron", "amy", "james", "yam" };

        Comparator<String> comparator = new Comparator<String>() {
            @Override public int compare(String o1, String o2) {
                System.out.println("Compare(" + o1 + "," + o2 + ")");
                char[] a1 = o1.toCharArray();
                Arrays.sort(a1);
                char[] a2 = o2.toCharArray();
                Arrays.sort(a2);
                return new String(a1).compareTo(new String(a2));
            }
        };
        Set<String> set = new TreeSet<String>(comparator);

        for (String name : name1) {
            set.add(name);
        }
        for (String name : name2) {
            set.add(name);
        }
        for (String name : name3) {
            set.add(name);
        }

        String[] result = set.toArray(new String[set.size()]);
        System.out.println(Arrays.asList(result));

        // Using TreeMap to keep the count.

        TreeMap<String, Integer> map = new TreeMap<String, Integer>(comparator);

        addAll(name1, map);
        addAll(name2, map);
        addAll(name3, map);

        System.out.println(map);
    }

    private static void addAll(String[] names, TreeMap<String, Integer> map) {
        for (String name : names) {
            if (map.containsKey(name)) {
                int n = map.get(name);
                map.put(name, n + 1);
            } else
                map.put(name, 1);
        }
    }
}

Answer 3

符合kdm：

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

public class RemoveDuplicateString {

    private static boolean add(Set<String> keySet, String s){
        char[] sortCharacters = s.toCharArray();
        Arrays.sort(sortCharacters);
        return keySet.add(new String(sortCharacters));
    }

    private static void check(Set<String> keySet, String []names, List<String> result){
        for (String name : names) {
            if (add(keySet, name)){
                result.add(name);
            }
        }
    }

    public static void main(String[] args) {
        String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"};
        String[] name2 = {"alan", "may", "jeremy", "helen", "alexi"};
        String[] name3 = {"adel", "aron", "amy", "james", "yam"};
        Set<String> keySet = new HashSet<String>();
        List<String> result = new ArrayList<String>();
        check(keySet, name1, result);
        check(keySet, name2, result);
        check(keySet, name3, result);
        System.out.println(result);
    }
}

Answer 4

另一种 Java 8 解决方案。

1) 使用规范化形式创建Map<String, List<String>，然后创建所有看到的不同形式

public static Map<String, List<String>> groupNormalised(final String[]... input) {
    return Arrays.stream(input)
            .flatMap(Arrays::stream)
            .collect(Collectors.groupingBy(s -> {
                char[] c = s.toCharArray();
                Arrays.sort(c);
                return new String(c);
            }));
}

例子：

Map<String, List<String>> grouped = groupNormalised(name1, name2, name3);        
grouped.forEach((k, v) -> System.out.printf("%s appears as %s%n", k, v));

输出：

eejmry appears as [jeremy, jeremy]
aceil appears as [alice]
eehln appears as [helen]
ejos appears as [jose]
adel appears as [adel]
aeilx appears as [alexi]
acikprt appears as [patrick]
aejms appears as [james]
amy appears as [amy, may, amy, yam]
anor appears as [aron]
aaln appears as [alan]

2）处理Map以提取您想要的数据

现在您有一个选择，您可以创建一个Set 的规范化形式：

final Set<String> normalisedForms = grouped.keySet();

或者您可以创建第一次相遇的Set：

final Set<String> first = grouped.values().stream()
        .map(c -> c.iterator().next())
        .collect(toSet());

或者作为一个数组：

final String[] first = grouped.values().stream()
        .map(c -> c.iterator().next())
        .toArray(String[]::new);