recursive function that checks if there are k words that they "charecter equale" to a given word

Question

we will say that two words are "charecter equale" if both of them has the same charecter, for example: baac and abac are charecter equale, I am trying to write a recursive function that gets a string s, a word w and integer k, that checks if there are exactliy k words in the string that they charecter equale to the word, for example: the function should return true for the word abac , the string aabc abdca caba xyz ab and the number k=2.

Ineed help at the recursive part, i.e the function searchMixed, my idea was first the check if the string contain only on word (base case),

the general case is to call the function searchMixed without the first word

public class recursion1 {


    public static void main(String[] args) {


        boolean result=searchMixed("abac","aabc abdca caba xyz ab",2);
        System.out.println("result: "+result);  
    }
    public static boolean searchMixed(String word, String s, int k)
    {

        if(s.indexOf(' ')==-1 && isEquale(word,s) && k==1)
            return true;

        if(s.indexOf(' ')==-1 && !isEquale(word,s) && k==1)
            return false;


        int pos=s.indexOf(' ');     
        System.out.println("index of"+ s.indexOf(' '));             
        String first_word=first_word=s.substring(0,pos);

        if(isEquale(word, first_word))
         searchMixed(word, s.substring(pos+1), k-1);
        else
         searchMixed(word, s.substring(pos+1), k);

        return false;
    }

.

    //this function works fine, the function checks if two words are charecter equale
    public static boolean isEquale(String word, String sub_string)
    {
        if(word.length()!=sub_string.length())
            return false;


        char[] s33=new char[sub_string.length()];
        char[] sww=new char[word.length()];


        for(int i=0;i<sub_string.length();i++)
            s33[i]=sub_string.charAt(i);        
        for(int i=0;i<word.length();i++)
            sww[i]=word.charAt(i);


        for(int i=0;i<word.length();i++)
        {
            for(int j=0;j<sub_string.length();j++)
            {
            if(sww[i]==s33[j])
                s33[j]='@';     
            }
        }

        for(int i=0;i<sub_string.length();i++)
            if(s33[i]!='@')
                return false;

        return true;

    }

}

Answer 1

YourString = YourString.replaceAll("\\s+", " ");
String[] words = YourString.split(" ");

...分开单词。

static int n = 0;
static String keyword = "aabc";
static String[] words = null;
public static void main()
{
    n = 0;
    // Let's assume you accept 'k' here.
    String YourString = "aabc baca hjfg gabac";
    words = YourString.split(" ");
    rec(words[0]);
    if (k <= n)
        System.out.println(true);
    else
        System.out.println(false);
}
static int pos = 0;
public static void rec(String word)
{
    boolean flag = true;

    word += " ";

    if(word.length() != keyword.length() + 1)
    {
        flag = false;
    }

    for(int i = 0; i < keyword.length() && flag; i++)
    {
        for(int j = 0; j < word.length(); j++)
        {
            if(word.charAt(j) == keyword.charAt(i))
            {
                word = word.substring(0, j) + word.substring(j+1);
                break;
            }
        }
        if(word.equals(" "))
        {
            n++;
            break;
        }
    }

    if(pos + 1 != words.length)
    {
        rec(words[++pos]);
    }
}

现在，让我解释一下：

1. 在递归方法rec(String word)中，在被检查单词的末尾添加一个空格（以便substring(j+1)不会超出范围）

   < / LI>

2. 如果关键字和选中的单词长度不同，则会停止检查，然后转到“5”。

3. 如果这两个单词的长度相同，那么循环会从单词中删除一个相似的字符（这就是word = word.substring(0, j) + word.substring(j+1);所做的）。

4. 在循环结束时，如果该单词的剩余部分是空格，则计数器n增加1并且循环退出。

5. 如果数组中有多于或等于一个字符串，则在数组中检查的字的位置增加1，并且数组中的下一个字传递给rec(String word)方法