[jzof] path in matrix 12

Typical DFS The problem of the algorithm ：

It is divided into four directions: up, down, left and right , To search .
Mark the search location before searching

import java.util.*;


public class Solution {
    
    public boolean hasPath(char[][] matrix, String word) {
    
        char[] words = word.toCharArray();
        for (int i = 0; i < matrix.length; i++) {
    
            for (int j = 0; j < matrix[0].length; j++) {
    
                // from [i,j] This coordinate starts to look up 
                if (dfs(matrix, words, i, j, 0))
                    return true;
            }
        }
        return false;
    }

    boolean dfs(char[][] matrix, char[] word, int i, int j, int index) {
    
        // The judgment of the boundary , If you cross the border and go back false.index It means to find a string word The first character of ,
        // If this character doesn't equal matrix[i][j], It shows that verifying this coordinate path is impossible , Go straight back to false
        if (i >= matrix.length || i < 0 || j >= matrix[0].length || j < 0 || matrix[i][j] != word[index])
            return false;
        // If word We've looked up every character of , Go straight back to true
        if (index == word.length - 1)
            return true;
        // Save the value of the current coordinate , In order to recover in the end 
        char tmp = matrix[i][j];
        // Then modify the value of the current coordinate 
        matrix[i][j] = '.';
        // Go recursion , Up, down, left, right along the current coordinates 4 Look in two directions 
        boolean res = dfs(matrix, word, i + 1, j, index + 1)
                || dfs(matrix, word, i - 1, j, index + 1)
                || dfs(matrix, word, i, j + 1, index + 1)
                || dfs(matrix, word, i, j - 1, index + 1);
        // Recursion and then restore the current coordinates 
        matrix[i][j] = tmp;
        return res;
    }
}