java - 查找java中单词排列的方式数

Question

我正在用 java 编写一个程序，它将计算一个单词可以排列/重新排列的方式数量。

例如，单词“HAPPY”可以以 60 种方式重新排列。

我的过程是找到单词长度的阶乘，然后将其除以单词中出现的不同字母的阶乘的乘积。(使用递归)。

这是我的代码:

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main
{
    public static int factorial(int fact)
    {
        if(fact == 0)
            return 1;
        else if(fact == 0)
            return 1;
        return(fact * (factorial(fact - 1)));

    }
    public static void main(String args[]) throws IOException
    {
        String word;
        int result;
        int cases,lengthOfWord;
        int a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z;
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    cases = Integer.parseInt(br.readLine());
    for(int iteration =1;iteration <= cases; iteration++)
    {
        word = br.readLine();
        lengthOfWord = word.length();
        a = lengthOfWord - word.replace("A", "").length();
        b = lengthOfWord - word.replace("B", "").length();
        c = lengthOfWord - word.replace("C", "").length();
        d = lengthOfWord - word.replace("D", "").length();
        e = lengthOfWord - word.replace("E", "").length();
        f = lengthOfWord - word.replace("F", "").length();
        g = lengthOfWord - word.replace("G", "").length();
        h = lengthOfWord - word.replace("H", "").length();
        i = lengthOfWord - word.replace("I", "").length();
        j = lengthOfWord - word.replace("J", "").length();
        k = lengthOfWord - word.replace("K", "").length();
        l = lengthOfWord - word.replace("L", "").length();
        m = lengthOfWord - word.replace("M", "").length();
        n = lengthOfWord - word.replace("N", "").length();
        o = lengthOfWord - word.replace("O", "").length();
        p = lengthOfWord - word.replace("P", "").length();
        q = lengthOfWord - word.replace("Q", "").length();
        r = lengthOfWord - word.replace("R", "").length();
        s = lengthOfWord - word.replace("S", "").length();
        t = lengthOfWord - word.replace("T", "").length();
        u = lengthOfWord - word.replace("U", "").length();
        v = lengthOfWord - word.replace("V", "").length();
        w = lengthOfWord - word.replace("W", "").length();
        x = lengthOfWord - word.replace("X", "").length();
        y = lengthOfWord - word.replace("Y", "").length();
        z = lengthOfWord - word.replace("Z", "").length();
        result = factorial(lengthOfWord) / (factorial(a)*factorial(b)*factorial(c)*factorial(d)*factorial(e)*factorial(f)*factorial(g)*factorial(h)*factorial(i)*factorial(j)*factorial(k)*factorial(l)*factorial(m)*factorial(n)*factorial(o)*factorial(p)*factorial(q)*factorial(r)*factorial(s)*factorial(t)*factorial(u)*factorial(v)*factorial(w)*factorial(x)*factorial(y)*factorial(z));
        System.out.printf("Data set %d: %d\n",iteration,result);
    }
}
}

但我认为它非常冗长且效率低下。

如何使这个程序更短、更高效？

我还想知道解决这个程序的其他方法。

请帮忙。谢谢。

Answer 1

您可以将字母计数保存在长度为 26 的数组中。此外，用于检查字符计数的字符串替换过于复杂。只需循环遍历字符并计算它们即可。 letters[word.charAt(j) - 'A'] 构造是一种诡计，它使 A 的计数出现在索引 0 处，B 出现在索引 1 处，等等。使用循环将阶乘相乘。并将单词转换为大写。最后，始终声明变量尽可能接近实际使用的位置。 (最后一个只是一般的良好实践。)

把它们放在一起:

public static void main(String args[]) throws IOException
{
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int cases = Integer.parseInt(br.readLine());
    for(int i = 0; i < cases; i++)
    {
        int letters = new int[26];
        String word = br.readLine().toUpperCase();
        int lengthOfWord = word.length();
        for (int j = 0; j < lengthOfWord; j++)
        {
            letters[word.charAt(j) - 'A']++;
        }
        int factorialProduct = 1;
        for (int j = 0; j < letters.length; j++)
        {
            factorialProduct *= factorial(letters[j]);
        }
        int result = factorial(lengthOfWord) / factorialProduct;
        System.out.printf("Data set %d: %d\n",iteration,result);
    }
}