使用基本Java检查两个字符串是否是彼此的字谜[重复]

Question

我正在 java Netbeans 中编写以下代码，这对于正常的字谜非常有效。但是，如果两个文本字段包含包含重复字母的单词，则代码将无法工作。可能是什么问题，我该如何解决？我对 Java 很基础，还不能理解数组。

String s1= t1.getText(); 
String s2= t2.getText();  
int b=0,c=0;
if(s1.length()!=s2.length())
   System.out.print("No");
else {
   for(int i=0;i<s1.length();i++) {
      char s = s1.charAt(i);
      for(int j=0;j<s2.length();j++) {
         if(s==s2.charAt(j)){
            b++;
         } 
      }
      if(b==0)
         break;
   }
   if(b==0)
      System.out.print("No");
   else 
      System.out.print("YES");
} 
System.out.print(b);

Answer 1

我会寻求更简单的推理：如果两个字符串在排序后完全匹配，则它们是字谜。 所以在Java中它会是这样的：

    String s1 = "cat";
    String s2 = "tac";
    boolean isAnagram = false;
    if (s1.length() == s2.length()) {
        char[] s1AsChar = s1.toCharArray();
        char[] s2AsChar = s2.toCharArray();
        Arrays.sort(s1AsChar);
        Arrays.sort(s2AsChar);
        isAnagram = Arrays.equals(s1AsChar, s2AsChar);
    } 

Answer 2

您想比较已排序的字符。这是一个单行：

return Arrays.equals(s1.chars().sorted().toArray(),
    s2.chars().sorted().toArray());

Arrays.equals() 为您比较长度和所有元素。

Answer 3

在我的解决方案中，我们计算第一个字符串中每个字符的出现次数，然后从第二个字符串中的计数中减去它。最后，检查字符数是否不为 0 则这两个字符串不是 anagram。

public static boolean isAnagram(String a, String b){
    //assume that we are using ASCII
    int[] charCnt = new int[256];
    for(int i = 0; i < a.length(); i++){
        charCnt[a.charAt(i)]++;
    }
    for(int i = 0; i< b.length(); i++){
        charCnt[b.charAt(i)]--;
    }
    for(int i = 0; i<charCnt.length; i++){
        if(charCnt[i] != 0) return false;
    }
    return true;
}

Answer 4

由于您似乎是初学者，因此这里有一个不涉及其他类或流中的函数的解决方案。它只涉及数组的使用以及char 也可以代表int 的事实。

public static void main(String[] args) throws ParseException {
    String s1= "anagram"; 
    String s2= "margana";  
    // We make use of the fact that a char does also represent an int.
    int lettersS1[] = new int[Character.MAX_VALUE];
    int lettersS2[] = new int[Character.MAX_VALUE];
    if(s1.length()!=s2.length())
       System.out.print("No");
    else {
       // Loop through the String once
       for(int i = 0; i<s1.length() ;++i) {
           // we can just use the char value as an index
           // and increase the value of it. This is our identifier how often 
           // each letter was aviable in the String. Alse case insensitive right now
           lettersS1[s1.toLowerCase().charAt(i)]++;
           lettersS2[s2.toLowerCase().charAt(i)]++;
       }
       // set a flag if the Strings were anagrams
       boolean anag = true;
       // We stop the loop as soon as we noticed they are not anagrams
       for(int i = 0;i<lettersS1.length&&anag;++i) {
           if(lettersS1[i] != lettersS2[i]) {
               // If the values differ they are not anagrams.
               anag = false;
           }
       }
       // Depending on the former loop we know if these two strings are anagrams
       if(anag) {
           System.out.print("Anagram");
       } else {
           System.out.print("No anagram");
       }
    } 
}

Answer 5

另一种解决方案，基于出现计数器：

static boolean areAnagrams(CharSequence a, CharSequence b) {
    int len = a.length();
    if (len != b.length())
        return false;

    // collect char occurrences in "a"
    Map<Character, Integer> occurrences = new HashMap<>(64);
    for (int i = 0; i < len; i++)
        occurrences.merge(a.charAt(i), 1, Integer::sum);

    // for each char in "b", look for matching occurrence
    for (int i = 0; i < len; i++) {
        char c = b.charAt(i);
        int cc = occurrences.getOrDefault(c, 0);
        if (cc == 0)                        
            return false;            
        occurrences.put(c, cc - 1);
    }
    return true;
}

虽然这个解决方案不如“排序和比较”那么优雅，但它可能更适用于不太可能成为字谜的长字符串，因为它在 O(n) 中运行而不是O(n logn) 并在第二个字符串的某个位置找不到匹配项时立即返回。

---

走出“Basic Java”领域，我修改了算法来处理surrogate pairs。这里收集和匹配的不是chars，而是int codepoints：

static boolean areAnagrams(CharSequence a, CharSequence b) {
    int len = a.length();
    if (len != b.length())
        return false;

    // collect codepoint occurrences in "a"
    Map<Integer, Integer> ocr = new HashMap<>(64);
    a.codePoints().forEach(c -> ocr.merge(c, 1, Integer::sum));

    // for each codepoint in "b", look for matching occurrence
    for (int i = 0, c = 0; i < len; i += Character.charCount(c)) {
        int cc = ocr.getOrDefault((c = Character.codePointAt(b, i)), 0);
        if (cc == 0)                        
            return false;            
        ocr.put(c, cc - 1);
    }
    return true;
}