剑指Offer-10矩阵中的路径

/**
 * 典型的回溯
 * @author Lucas
 * @date 2021/02/07 22:49
 * @param board 二维数组字典
 * @param word 单词
 * @return boolean
 */
public static boolean exist(char[][] board, String word) {

    int h = board.length;
    int w = board[0].length;
    //经过矩阵的某个格子不允许被重复使用，visited用于标记是否已经经过
    boolean[][] visited = new boolean[h][w];
    //因为每一个格子都可以作为起始点，所以两次循环不可避免了
    for (int i = 0; i < h; i++){
        for (int j = 0; j < w; j++){
            //解释一下index用户标记当前匹配到word中的第几个字符
            boolean res = search(board, word, i, j, 0, visited);
            if (res){
                return true;
            }
        }
    }
    return false;
}

public static boolean search(char[][] board, String word, int i, int j, int index, boolean[][] visited){
    //word已经全部匹配成功
    if (index >= word.length()){
        return true;
    }
    //i j越界、已经访问过了、word当前字符与board中的不匹配
    if (i < 0 || j < 0 || i >= board.length || j>= board[0].length || visited[i][j] || board[i][j] != word.charAt(index)){
        return false;
    }
    //经过了的记为true
    visited[i][j] = true;
    //上下左右四个方向 只要有一个方向匹配即可
    boolean res = search(board, word , i + 1, j, index + 1, visited) || search(board, word , i - 1, j, index + 1, visited) || search(board, word , i, j + 1, index + 1, visited) || search(board, word , i, j - 1, index + 1, visited);
    //进入到下一个循环前，重置visited为false
    visited[i][j] = false;
    return res;
}