剑指 Offer 12. 矩阵中的路径

package com.example.lettcode.offer;

/**
 * @Class Exist
 * @Description 剑指 Offer 12. 矩阵中的路径
 * 请设计一个函数，用来判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。
 * 路径可以从矩阵中的任意一格开始，每一步可以在矩阵中向左、右、上、下移动一格。
 * 如果一条路径经过了矩阵的某一格，那么该路径不能再次进入该格子。
 * 例如，在下面的3×4的矩阵中包含一条字符串“bfce”的路径（路径中的字母用加粗标出）。
 * <p>
 * [["a","b","c","e"],
 * ["s","f","c","s"],
 * ["a","d","e","e"]]
 * 但矩阵中不包含字符串“abfb”的路径，因为字符串的第一个字符b占据了矩阵中的第一行第二个格子之后，
 * 路径不能再次进入这个格子。
 * <p>
 * 示例 1：
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
 * 输出：true
 * <p>
 * 示例 2：
 * 输入：board = [["a","b"],["c","d"]], word = "abcd"
 * 输出：false
 * 提示：
 * <p>
 * 1 <= board.length <= 200
 * 1 <= board[i].length <= 200
 * @Author
 * @Date 2020/6/29
 **/
public class Exist {
    /**
     * 解法1:利用dfs + 剪枝法
     */
    public static boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0) return false;
        char[] words = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (dfs(board, words, i, j, 0)) return true;
            }
        }
        return false;
    }

    private static boolean dfs(char[][] board, char[] words, int i, int j, int k) {
        // 终止条件 越界或者全部匹配
        // 其中k表示words中第k个匹配的字符
        if (i >= board.length || i < 0
                || j >= board[0].length || j < 0
                || k>=words.length || board[i][j] != words[k])
            return false;
        // 表示words全部匹配
        if(k == words.length-1) return true;

        // 标记当前元素，防止重复搜索
        char temp = board[i][j];
        board[i][j] = '/';
        // 上下左右四个方向
        boolean flag = dfs(board, words, i + 1, j, k + 1) || dfs(board, words, i, j + 1, k + 1)
                || dfs(board, words, i - 1, j, k + 1) || dfs(board, words, i, j - 1, k + 1);

        // 元素还原
        board[i][j] = temp;
        return flag;
    }

    public static void main(String[] args) {
        char[][] board = new char[][]{
                {'a', 'b', 'c', 'e'},
                {'s', 'f', 'c', 's'},
                {'a', 'd', 'e', 'e'}};
        String word = "abcced";
        boolean ans = exist(board, word);
        System.out.println("demo01 result:" + ans);

        board = new char[][]{{'a', 'b'}, {'c', 'd'}};
        word = "abcd";

        ans = exist(board, word);
        System.out.println("demo02 result:" + ans);
    }
}