79. Word Search (Medium) Facebook Microsoft Bloomberg

Given a 2D board and a word, find if the word exists in the grid.

The word can be constructed from letters of sequentially adjacent cell, where "adjacent" cells are those horizontally or vertically neighboring. The same letter cell may not be used more than once.

For example,
Given board =

[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

word = "ABCCED", -> returns true,

word = "SEE", -> returns true,

word = "ABCB", -> returns false.

Solution 1: DFS O(m * n * 4^k); O(k) k is the word's length.

time complexity 似乎是O(mn*3^len) ?因为除了最开始有4个方向的选择，以后都是只有3个方向的选择（意思是单词不能往回走。）

    /**
     * DFS O(mn*4^k); O(k)
     * For each character c in board,
     *   Start DFS if c matches the first character of string.
     */
    public boolean exist(char[][] board, String word) {
        if (board == null || board.length == 0 || word == null) {
            return false;
        }
        if (word.length() == 0) {
            return true;
        }

        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == word.charAt(0)) {
                    if (dfs(board, i, j, word, 0)) {
                        return true;
                    }
                }
            }
        }
        return false;
    }

    private boolean dfs(char[][] board, int i, int j, String word, int start) {
        if (start == word.length()) {
            return true;
        }

        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != word.charAt(start)) {
            return false;
        }

        board[i][j] = '#';
        boolean res = dfs(board, i + 1, j, word, start + 1)
                || dfs(board, i - 1, j, word, start + 1)
                || dfs(board, i, j + 1, word, start + 1)
                || dfs(board, i, j - 1, word, start + 1);
        board[i][j] = word.charAt(start); //backtrack
        return res;
    }

Follow Up:

1.不能在原始char[][]做标记，于是把visited boolean[][] 换成了stack<position>.
http://www.1point3acres.com/bbs/thread-297336-1-1.html