我现在正面临着一个问题。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个数组,例如,
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像may
、amy
和yam
这样的字符串。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。也就是说,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和仅包括amy)。
到目前为止,我使用HashSet
进行的尝试如下所示
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,这段代码的问题是,它只是删除了同一字符串的多次出现。还有没有别的选择?任何帮助都是非常感谢的。
发布于 2015-06-23 15:03:09
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);
}
}
}
发布于 2015-06-23 14:44:05
您可以对字符串的字符数组(str.toCharArray ()
)进行排序,并从排序后的数组创建一个新字符串,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
中,并检查每个字符串的规范表示是否已经存在于集合中。
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>
,因为数组不覆盖hashCode
和equals
,所以HashSet<char[]>
不起作用。
发布于 2015-06-23 15:58:21
与kdm一致:
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);
}
}
https://stackoverflow.com/questions/30995553
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