这些是我的项目的详细信息:
Anagrams:该项目的目的是创建一款向用户呈现的游戏
一个单词的字谜,必须在有限的时间内猜出正确的单词
尝试。
项目特色:
我面临的问题:
我能够从字符串数组中取出一个随机单词,但无法将其与输出进行比较,因为输出是普通字符串。我想知道如何解决这个问题以及应该如何进一步进行,我不需要提示部分的答案,只是想知道如何比较两种类型的字符串。
import java.util.Arrays;
import java.io.ByteArrayOutputStream;
import java.io.PrintStream;
import java.io.*;
import java.util.Scanner;
import java.util.*;
public class main {
String A = "words[index]";
static boolean isAnagram(String A, String B) {
if (A.length() != B.length()) {
return false;
} else {
char a[] = A.toLowerCase().toCharArray();
char b[] = A.toLowerCase().toCharArray();
Arrays.sort(a);
Arrays.sort(b);
String sortedA = String.valueOf(a);
String sortedB = String.valueOf(B);
if (sortedA.equals(sortedB)) {
}
}
return false;
}
public static void main(String[] args) {
String[] words = {"Rat", "Car", "Below", "Taste", "Cried", "Study", "Thing", "Chin"};
Random random = new Random();
int index = random.nextInt(words.length);
System.out.println("The given word is: " + words[index]);
Scanner sc = new Scanner(System.in);
String B = sc.next();
if (isAnagram(String A, String B)) {
System.out.println("not an angram");
} else {
System.out.println("Sucess");
}
}
}
您可以从这些字符串中获取两个字符数组,对它们进行排序,然后检查这些数组是否彼此相等。
/**
* @param a first string.
* @param b second string.
* @return whether two strings are case insensitive anagrams of each other.
*/
static boolean isAnagram(String a, String b) {
// invalid incoming data
if (a == null || b == null
|| a.length() != b.length()) return false;
return Arrays.equals(
a.toLowerCase().codePoints().sorted().toArray(),
b.toLowerCase().codePoints().sorted().toArray());
}
public static void main(String[] args) {
System.out.println(isAnagram("Study", "dusty")); // true
System.out.println(isAnagram("door", "rooD")); // true
System.out.println(isAnagram("door", "root")); // false
}
我可以提供两种方法来检查两个字符串是否是字谜:
1.
static boolean isAnagram(String A, String B) {
if (A.length() != B.length()) return false;
else {
char[] a = A.toLowerCase().toCharArray();
char[] b = A.toLowerCase().toCharArray();
Arrays.sort(a);
Arrays.sort(b);
for (int i = 0; i < a.length; i++)
if (a[i] != b[i]) return false;
return true; //time Complexity: O(nLogn)
}
}
static int NUM_OF_CHARS = 256;
static boolean isAnagram(String A, String B) {
if (A.length() != B.length()) return false;
else {
int[] count1 = new int[NUM_OF_CHARS];
Arrays.fill(count1, 0);
int[] count2 = new int[NUM_OF_CHARS];
Arrays.fill(count2, 0);
char[] a = A.toCharArray();
char[] b = A.toCharArray();
for (int i = 0; i < a.length && i < b.length; i++) {
count1[a[i]]++;
count2[b[i]]++;
}
for (int i = 0; i < NUM_OF_CHARS; i++)
if (count1[i] != count2[i]) return false;
return true;
}//time Complexity: O(n)
}
使用集合检查两个字符串是否是字谜:
package StringPackage;
import java.util.*;
public class AnagramUsingCollections {
public static void main(String[] args) {
// TODO Auto-generated method stub
char[] string1 = "aab".toCharArray();
char[] string2 = "baa".toCharArray();
Map<Character, Integer> hm = new HashMap<>();
for (char no : string1) {
//for start
Integer count = hm.get(no);
if (count == null) {
hm.put(no, 1);
} else {
count++;
hm.put(no, count);
}
//for end
}
if (string1.length == string2.length) {
for (char no : string2) {
//baa
Integer count = hm.get(no);
if (count == null) {
hm.put(no, 1);
} else {
count--;
hm.put(no, count);
}
}//for end
}
Set<Map.Entry<Character, Integer>> se = hm.entrySet();
for (Map.Entry<Character, Integer> me : se) {
if (me.getValue() > 0) {
System.out.println("Not an anagram");
break;
} else {
System.out.println("IS an anagram");
break;
}
}
}
}
var isAnagram = function(s, t) {
if (s.length !== t.length) {
return false; // Different lengths cannot be anagrams
}
const sCount = {};
const tCount = {};
s.split("").forEach(letter => {
sCount[letter] = (sCount[letter] || 0) + 1;
});
t.split("").forEach(letter => {
tCount[letter] = (tCount[letter] || 0) + 1;
});
// Check if all keys and values in tCount are present in sCount
for (const key in tCount) {
if (!(key in sCount) || sCount[key] !== tCount[key]) {
return false; // Found a key that is missing or has a different count
}
}
return true; // All keys and values are the same in both objects
};