79. Word Search

原题链接：https://leetcode.com/problems/word-search/description/
做完上一题：212. Word Search II，我就知道有二必有一，那么再做这第一道题目就太简单了，不废话思路是一样的还是回溯法：

class Solution {
    public boolean exist(char[][] board, String word) {
        char[] wordChars = word.toCharArray();
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (dfs(board, i, j, wordChars, 0)) {
                    return true;
                }
            }
        }
        return false;
    }
    
    public boolean dfs(char[][] board, int i, int j, char[] wordChars, int k) {
        char c = board[i][j];
        if (c == '#' || wordChars[k] != c) {
            return false;
        }
        k++;
        if (k == wordChars.length) {
            return true;
        }

        board[i][j] = '#';
        if (i > 0) {
            if (dfs(board, i - 1, j, wordChars, k)) {
                return true;
            }
        }
        if (j > 0) {
            if (dfs(board, i, j - 1, wordChars, k)) {
                return true;
            }
        }
        if (i < board.length - 1) {
            if (dfs(board, i + 1, j, wordChars, k)) {
                return true;
            }
        }
        if (j < board[0].length - 1) {
            if (dfs(board, i, j + 1, wordChars, k)) {
                return true;
            }
        }
        board[i][j] = c;
        return false;
    }
}