c - 在二维字符数组中搜索单词

Question

给定一个大小为 100 x100 的字符的 2D 数组和一个单词（1D 字符数组），在 2D 数组中找到给定单词的出现次数（仅从左到右水平搜索）。

char data[100][100] =
{
    "ACFRTBOOK",
    "BOPNBOOKQUIZGEEK",
    "IDEQAPRACTICE"
};
char  Word[4] = "BOOK";

输出：

pattern found at 0, 5
pattern found at 1, 4

Answer 1

首先，您不能像使用char Word[4] = "BOOK";此操作那样将 char 数组初始化为 String，这必须char word[4] = {'b','o','o','k'};同样适用于您的 2D 数组命名数据。

所以这就是你将如何做这样的事情。我在//评论中添加了解释，但总体思路是将数组转换为字符串，因为在定位另一组字符时更容易使用。

如果您需要进一步的帮助，请告诉我。我希望这有帮助。

public void findWord(char[] word; char[][] data){
    String w = new String(word); //it is much easier to work with strings for a searching problem like this
    int x=0,y=0;
    for(char[] i:data){//iterates through each row of data
        String d = new String(i);
        while(d.contains(w)){
            x+=i.indexOf(w);//locates the pattern
            System.out.println("pattern found at " + x + ", " + y);
            x++;
            d=s.substring(x);//this removes the previous occurance of the patern so that the index of can find a new repeat in the same row 
        }
        y++;
        x=0;//reset for next array
    }
}

Answer 2

这可能会有所帮助：

int k = 0, n = 0;
char word[] = "BOOK";

for (int i = 0; i < size - 1; i++)
{
    for (int j = 0; data[i][j] != '\0'; j++)
    {
        n = j;

        while (data[i][n] == word[k] && word[k] != '\0')
        {
            n++;
            k++;

            if (word[k] == '\0')
            {
                printf("Found at %i, %i", i, j)
            }
        }

        k = 0;
    }
}