Question

我有一个字符矩阵，假设是4 * 5矩阵

A-C-P-的˚F

X-S-的 0 -P

V-的û -Q-N

W-G-的Ñ -M

d -A-T-I

我想知道该矩阵中是否存在 FOUND 字样。

我的矩阵如下（JAVA代码）

char matrixOfChars[][]={
{'A','C','P','F'},
{'X','S','O','P'},
{'V','U','Q','N'},
{'W','G','N','M'},
{'D','A','T','I'}};

注意：在矩阵中，第一个字总是在第一行，第二个字总是在第二行，依此类推

目前为单个字符串完成的代码在

之下

private static void generateCombinations(String original, String combination) {
        if (original.length() == 0) {
            System.out.println(combination);
        } else {
            for(int i=0;i<original.length();i++){
                generateCombinations(original.substring(0,i) + original.substring(i+1,original.length()),combination + original.charAt(i));
            }
        }
    }

请帮我在JAVA中找到上面的解决方案。如果使用java递归给出解决方案会更好。

谢谢。

Answer 1

使用循环迭代FOUND的字符，在矩阵的每一行中使用Arrays.asList和indexOf来获取位置并创建一个新数组或所需内容

String word = "FOUND";
int[] sollution = new int[work.length()];

for (int i = 0; i < word.length(); i++) {
    int index = Arrays.asList(matrixOfChars[i]).indexOf(word.charAt(i));
    System.out.print(index + "\n");       
    sollution[i] = index;
}

Arrays.toString(sollution);

您将获得此输出：

sollution = {3 , 2 , 1 , 2 , 0 }

Answer 2

试试此代码

public class Test {


public static void main(String[] args) {
    char matrixOfChars[][]={
            {'A','C','P','F'},
            {'X','S','O','P'},
            {'V','U','Q','N'},
            {'W','G','N','M'},
            {'D','A','T','I'}};

    System.out.println(find(matrixOfChars,5,4, "FOUND"));
}

public static boolean find(char matrixOfChars[][],int rowsSize,int columnSize,String word)
{
    if(word.length()!=rowsSize)
        return false;
    String temp="";
    for (int i = 0; i < rowsSize; i++) {
        for (int j = 0; j < columnSize; j++) {
            if(word.charAt(i)==matrixOfChars[i][j])
            {
                temp+=word.charAt(i);
                break;
            }
        }
    }
    System.out.println(temp);
    if(temp.equals(word))
    {
        return true;
    }
    else
    {
        return false;
    }
}


}

递归解决方案：

public static void main(String[] args) {
    char matrixOfChars[][] = { { 'A', 'C', 'P', 'F' },
            { 'X', 'S', 'O', 'P' }, { 'V', 'U', 'Q', 'N' },
            { 'W', 'G', 'N', 'M' }, { 'D', 'A', 'T', 'I' } };
    String word="AGUNC";
    String res=find2(matrixOfChars, 5, 4, word);
    System.out.println(res);
    System.out.println(res.equals(word));

}

static String temp = "";
static int n = 0;

public static String find2(char matrixOfChars[][], int rowsSize,
        int columnSize, String word) {

    if (word.length() != rowsSize || n == rowsSize)
        return temp;
    else {

        for (int j = 0; j < columnSize; j++) {
            if (word.charAt(n) == matrixOfChars[n][j]) {
                temp += word.charAt(n);
                n++;
                return find2(matrixOfChars, rowsSize, columnSize, word);
            }

        }
        n++;
        return "";
    }
}

Answer 3

使用递归：

class Test {

    static char matrixOfChars[][]={
            {'A','C','P','F'},
            {'X','S','O','P'},
            {'V','U','Q','N'},
            {'W','G','N','M'},
            {'D','A','T','I'}};

    static int n = 0;

    public static void main(String args[]) {
        n = matrixOfChars.length - 1;
        find();
    }

    public static void find() {
        String word = "";

        if(n >= 0) {

            for(int i = 0 ; i < matrixOfChars[n].length ; i ++) {
                word += matrixOfChars[n][i];
            }

            if(word.equals("FOUND")) {
                System.out.println("FOUND is found");

            } else {

                n--;
                find();
            }

        } else {
            System.out.println("FOUND is not found");
        }

    }
}

Answer 4

尝试以下代码：

    char matrixOfChars[][]={
    {'A','C','P','F'},
    {'X','S','O','P'},
    {'V','U','Q','N'},
    {'W','G','N','M'},
    {'D','A','T','I'}};

    String word = "FOUND";
    int[] sol = new int[word.length()];

    for (int i = 0; i < word.length(); i++) {

        char y = word.charAt( i );
        int z = new String( matrixOfChars[i] ).indexOf( y );

        System.out.print( z + "\n" );
        sol[i] = z;
    }

    System.out.println( Arrays.toString( sol ) );

在字符矩阵中，找到特定的字符串

4 个答案: