[英]Java using hashmap for searching string
Is it possible to compare characters from a string and print out the position of the first unique one?是否可以比较字符串中的字符并打印出第一个唯一字符的 position? Is there maybe something from the String class that can help me do this?字符串 class 是否有一些东西可以帮助我做到这一点?
Pseudocode:伪代码:
enter code here
String s = "ABC456";
int n = 2;
ArrayList<String> str = new ArrayList<>();
Map<String, Long> counts2 = new LinkedHashMap<String,
Long>();
for(String val : str){
long count = counts2.getOrDefault(val, 0L);
counts2.put(val, ++count);
}
for(String key: counts2.keySet()){
if(counts2.get(key)==1){
System.out.println(list.indexOf(key));
break;
}
}
Please try mine:请试试我的:
import java.util.HashSet;
import java.util.HashMap;
import java.util.LinkedHashSet;
public class Main {
public static void main(String[] args) {
String originalString = "AAAB"; //little trickier input! ;)
int n = 1;
LinkedHashSet<String> uniques = new LinkedHashSet<>();
HashSet<String> dupes = new HashSet<>();
HashMap<String, Integer> str2Idx = new HashMap<>();
for (int cur = 0; cur <= originalString.length() - n; cur++) {
String substr = originalString.substring(cur, cur + n);
if (uniques.contains(substr)) { // cleanup
uniques.remove(substr);
str2Idx.remove(substr);
dupes.add(substr);
} else if(!dupes.contains(substr)){ // store
uniques.add(substr);
str2Idx.put(substr, cur);
}
}
// when the input is "external"/unknown, we should also avoid:
if (uniques.isEmpty()) {
System.out.println(-1);
} else {
// Print(get the index of (first element of uniques))
System.out.println(str2Idx.get(uniques.iterator().next()));
// Prints "3" with the given input
}
}
}
So basically:所以基本上:
a LinkedHashSet
for unique substrings.用于唯一子字符串的LinkedHashSet
。
contains
operation faster和“哈希”:使contains
操作更快a HashMap<String, Integer
, as the (variable) name suggest:一个HashMap<String, Integer
,正如(变量)名称所暗示的:
an additional "dupes" storage, to avoid re-adding.额外的“重复”存储,以避免重新添加。
Please test it deeper.请更深入地测试它。
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