对anagram的改进

Question

刚刚完成了最近的作业，但我知道它可能会更有效率。它从命令行中读取两个单词，忽略空格和标点符号，并确定它们是否是变形词。我所拥有的是下面的；据我所知，它是全功能的。

/**
 * Find out if a string is an anagram of another string
 * 
 */

import java.util.Arrays;

public class Anagram
{
    public static void main(String[] args)
    {
        if (args.length != 2)
            System.out.println("You did not enter two words!");        
        else            
            printWords(args[0], args[1]);                              
    }

    // method to determine whether two strings have the same chars
    public static boolean areAnagrams(String wordOne, String wordTwo) 
    {
        // new strings for letters only
        String ltrsOnlyOne = lettersOnly(wordOne);
        String ltrsOnlyTwo = lettersOnly(wordTwo);      

        // convert strings to lowercase char arrays
        char[] first = ltrsOnlyOne.toLowerCase().toCharArray();
        char[] second = ltrsOnlyTwo.toLowerCase().toCharArray();

        // sort char arrays using sort method
        Arrays.sort(first);
        Arrays.sort(second);

        if (Arrays.equals(first, second))
            return true;
        else
            return false;
    }

    public static String lettersOnly(String word) 
    {
        int length = word.length();
        StringBuilder end = new StringBuilder(length);
        char x;

        for (int i = (length - 1); i >= 0; i--) {
            x = word.charAt(i);
            if (Character.isLetter(x)) {
                end.append(x);
            }
        }
        return end.toString();
    }

    public static void printWords(String wordOne, String wordTwo)
    {
       boolean b = areAnagrams(wordOne, wordTwo);
       if (b == true) {
            System.out.println(wordOne + " is an anagram of "+ wordTwo);
       }

       if (b == false) {
            System.out.println(wordOne + " is not an anagram of "+ wordTwo);
       }
    }
}

Answer 1

• Execution time:如果您想提高执行时间，可以统计每个字母在每个单词中出现的次数，然后进行比较。这样，您可以避免排序(具有O(n * log n)复杂性)，并在线性(O(n))时间内解决问题。
• javadoc :一个次要问题是，类(/** ... */)的javadoc注释应该紧跟在类声明(public class ...)之前。现在，在javadoc注释和类声明之间有一条导入语句。

Answer 2

我会将santize步骤lettersOnly()移到sameChars()之外的一个单独的方法中，从main调用。它不是更高效，但它的代码更干净：

public static void main(String[] args) 
{ 
    if (args.length != 2) { 
        System.out.println("You did not enter two words!"); 
    } 

    String wordOne = lettersOnly(args[0]); 
    String wordTwo = lettersOnly(args[1]);    

    printWords(wordOne, wordTwo);              
} 

// method to determine whether two strings have the same chars 
public static boolean sameChars(String wordOne, String wordTwo)  
{ 
    // convert strings to lowercase char arrays 
    char[] first = wordOne.toLowerCase().toCharArray(); 
    char[] second = wordTwo.toLowerCase().toCharArray(); 

    Arrays.sort(first); 
    Arrays.sort(second); 

    return  Arrays.equals(first, second);
} 

public static void printWords(String wordOne, String wordTwo) 
{ 
    boolean isAnagram = sameChars(wordOne, wordTwo);         

   if (isAnagram) 
   { 
        System.out.println(wordOne + " is an anagram of "+ wordTwo); 
   } 
   else
   { 
        System.out.println(wordOne + " is not an anagram of "+ wordTwo); 
   } 
} 

Answer 3

您只需要调用sameChars一次。否则，在false情况下，您将小写并迭代数组两次。尝试：

public static void printWords(String wordOne, String wordTwo)
 {
   boolean same = sameChars(wordOne, wordTwo);
   if (same) {
        System.out.println(wordOne + " is an anagram of "+ wordTwo);
    } else {
        System.out.println(wordOne + " is not an anagram of "+ wordTwo);
    }
}