Question

我正在研究leetcode problem。我想出了一个简单的解决方案，但它给出了错误的输出。

给出2D板和一个单词，查找单词是否存在于网格中。

单词可以由顺序相邻的字母构成   单元格，其中“相邻”单元格是水平或垂直的单元格   邻接。同一字母单元不得重复使用。

对于此输入：

面板：[[“ a”，“ b”]，[“ c”，“ d”]]

单词：“ abcd”

它应该返回false，但是下面的解决方案将返回true。

public static boolean exist(char[][] board, String word) {
    int row = board.length;
    int col = board[0].length;
    Map<Character, Integer> hm = new HashMap<Character, Integer>();
    for (int i = 0; i < row; i++) {
      for (int j = 0; j < col; j++) {
        hm.put(board[i][j], hm.getOrDefault(board[i][j], 0) + 1);
      }
    }

    char[] words = word.toCharArray();
    for (int i = 0; i < words.length; i++) {
      char x = words[i];
      if (hm.containsKey(x) && hm.get(x) > 0)
        hm.put(x, hm.get(x) - 1);
      else
        return false;
    }
    return true;
  }

Answer 1

使用DFS（深度优先搜索）算法：

public boolean exist(char[][] board, String word) {
    int m = board.length;
    int n = board[0].length;

    boolean result = false;
    for(int i=0; i<m; i++){
        for(int j=0; j<n; j++){
           if(dfs(board,word,i,j,0)){
               result = true;
           }
        }
    }

    return result;
}

public boolean dfs(char[][] board, String word, int i, int j, int k){
    int m = board.length;
    int n = board[0].length;

    if(i<0 || j<0 || i>=m || j>=n){
        return false;
    }

    if(board[i][j] == word.charAt(k)){
        char temp = board[i][j];
        board[i][j]='#';
        if(k==word.length()-1){
            return true;
        }else if(dfs(board, word, i-1, j, k+1)
        ||dfs(board, word, i+1, j, k+1)
        ||dfs(board, word, i, j-1, k+1)
        ||dfs(board, word, i, j+1, k+1)){
            return true;
        }
        board[i][j]=temp;
    }

    return false;
}

Answer 2

这是使用Java流的另一种方法。希望有帮助！

import java.util.Arrays;
import java.util.stream.IntStream;

public class WordSearch{
     public static void main(String []args){
        final char[][] matrix = {
            {'X', 'H', 'A', 'T'},
            {'X', 'E', 'X', 'X'},
            {'X', 'L', 'X', 'O'},
            {'X', 'L', 'X', 'W'},
            {'C', 'O', 'O', 'L'}
        };

        previewMatrix(matrix);

        String[] searchSequences = {"COOL", "HAT", "HELLO", "OWL", "WORLD"};

        for (String searchSequence : searchSequences) {
            boolean isWordFound = searchWord(searchSequence, matrix);

            if (isWordFound) {
                System.out.printf("[ PASS ] The character sequence %s has been found in the matrix.\n", searchSequence);
            } else {
                System.out.printf("[ FAIL ] The character sequence %s was not found in the matrix.\n", searchSequence);
            }
        }
     }

     private static boolean searchWord(String searchSequence, char[][] matrix) {
         boolean isWordFound = false;
         String rowString = null;

         for (char[] row : matrix) {
             rowString = new String(row);

             if (rowString.contains(searchSequence)) {
                   isWordFound = true;
                   break;
             }
         }

         if (!isWordFound) {
             int columnsCount = matrix[0].length;
             String colString = null;

             for (int colIndex=0; colIndex < columnsCount; colIndex++) {
                   colString = getColumnContent(matrix, colIndex); 
                   if (colString.contains(searchSequence)) {
                        isWordFound = true;
                        break;
                   }
             }
         }

         return isWordFound;
     }

     private static void previewMatrix(char[][] matrix) {
         for (char[] row : matrix) {
            for (char ch : row) {
                System.out.print(ch + " ");
            }

            System.out.println();
        }

        System.out.println();
     }

     private static String getColumnContent(char[][] matrix, int columnIndex) {
         return IntStream
            .range(0, matrix.length)
            .mapToObj(i -> (char) matrix[i][columnIndex])
            .collect(
                StringBuilder::new,
                StringBuilder::appendCodePoint,
                StringBuilder::append)
            .toString();
     }
}

在字符网格中搜索单词

2 个答案: