[英]How to search String array through char array?
我想構建一個不使用字典庫的簡單字典搜索程序。 如果數組中的字符串等於char數組,我想搜索字符串數組。 它應該打印字符串及其索引號。 如果有兩個與char匹配的字符串,則應打印第一個字符串,並保留后一個字符串
.eg字符串array [“ fine”,“ rest”,“ door”,“ shine”] char字符['t','r','e','s']。 答案應該是索引1處的“ rest”。如果重復了String rest,那么它應該只打印第一個。
我試圖比較字符串數組和char數組,但是它返回匹配char數組的所有字符串單詞。
String strArray[]=new String[4];
char chrArray[]=new char[4];
String value="";
char compare;
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
for (int i = 0; i < strArray.length; i++) {
if(strArray[i].length()==chrArray.length){
if(""+strArray[i]!=value){
value="";
}
for (int j = 0; j < strArray[i].length(); j++) {
for (int k = 0; k < chrArray.length; k++) {
if(strArray[i].charAt(j)==chrArray[k]){
value=value+strArray[i].charAt(j);
}
}
}
}
}
System.out.println(value);
輸出應該是來自數組的字符串,該字符串等於char數組。
您可以對char數組進行排序,然后使用Arrays.equal
對其進行比較。 通過對char數組進行排序,將不需要為循環使用2。
import java.util.Arrays;
import java.util.Scanner;
public class Bug {
public static void main(String[] args) {
String strArray[]=new String[4];
char chrArray[]=new char[4];
Scanner input = new Scanner(System.in);
System.out.println("Enter the words :");
for(int i=0;i<strArray.length;i++){
strArray[i]=input.next();
}
System.out.println("Enter the Characters :");
for (int i = 0; i < chrArray.length; i++) {
chrArray[i]=input.next().charAt(0);
}
Arrays.sort(chrArray);
for (int i = 0; i < strArray.length; i++) {
char[] x = strArray[i].toCharArray();
Arrays.sort(x);
if(Arrays.equals(chrArray, x))
{
System.out.println(strArray[i]);
break;
}
}
}
}
您還可以創建Map<Character, Integer>
來存儲單詞和字符的計數,並將每個單詞的字符計數與字符計數進行比較。 如果兩者相等,則將其打印出來。
演示:
import java.util.Map;
import java.util.HashMap;
public class Example {
private static final String[] words = {"fine", "rest", "door", "shine"};
private static final char[] chars = {'t', 'r', 'e', 's'};
/**
* Initialise String character counts in a hashmap datastructure.
* @param word The string word to iterate.
* @return Hashmap of character -> counts.
*/
private static Map<Character, Integer> getCharCounts(String word) {
Map<Character, Integer> wordCounts = new HashMap<>();
for (int i = 0; i < word.length(); i++) {
Character character = word.charAt(i);
// If key doesn't exist, initialise it
if (!wordCounts.containsKey(character)) {
wordCounts.put(character, 0);
}
// Increment count by 1
wordCounts.put(character, wordCounts.get(character) + 1);
}
return wordCounts;
}
public static void main(String[] args) {
// Initialise character counts first
Map<Character, Integer> charCounts = getCharCounts(String.valueOf(chars));
for (int i = 0; i < words.length; i++) {
Map<Character, Integer> wordCounts = getCharCounts(words[i]);
// If Hashmaps are equal, then we have found a match
if (wordCounts.equals(charCounts)) {
System.out.println(words[i] + ", at index = " + i);
break;
}
}
}
}
輸出:
rest, at index = 1
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