内容来源于 Stack Overflow,并遵循CC BY-SA 3.0许可协议进行翻译与使用
我有三个数组,
String[] name1 = {"amy", "jose", "jeremy", "alice", "patrick"}; String[] name2 = {"alan", "may", "jeremy", "helen", "alexi"}; String[] name3 = {"adel", "aron", "amy", "james", "yam"};
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。最后的数组应该是:
String[] finalArray={"amy", "jose", "alice", "patrick","alan", "jeremy", "helen", "alexi","adel", "aron", "james"}
我试过HashSet
,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。还有别的选择吗?
你可以对String(str.toCharArray ()
)的字符数组进行排序,并从排序的数组中创建一个新的String以获取String的“规范”表示形式。
然后,可以将这些字符串添加到一个Set
,并检查每个字符串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));
}