[Leetcode]Word Search

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 =

[  ["ABCE"],  ["SFCS"],  ["ADEE"]]

word = "ABCCED", -> returns true,
word = "SEE", -> returns true,

word = "ABCB", -> returns false.

在矩阵中找是否存在某个单词~用DFS来解~先在矩阵board中找到word的第一个字母，然后从这个元素出发，往上下左右深度搜索是否有相等于word的字符串~注意要维护一个数组used来标记元素是否已经被访问过了

class Solution:    # @param board, a list of lists of 1 length string    # @param word, a string    # @return a boolean    def exist(self, board, word):        if board is None or len(board) == 0 or len(board[0]) == 0: return False        used = [[False] * len(board[0]) for i in xrange(len(board))]        for i in xrange(len(board)):            for j in xrange(len(board[0])):                if board[i][j] == word[0]:                    if self.helper(board, word, 0, i, j, used):                        return True        return False            def helper(self, board, word, index, i, j, used):        if index == len(word):            return True        if i < 0 or j < 0 or i >= len(board) or j >= len(board[0]) or used[i][j] or word[index] != board[i][j]:            return False        used[i][j] = True        res = self.helper(board, word, index + 1, i + 1, j, used) or self.helper(board, word, index + 1, i - 1, j, used) or self.helper(board, word, index + 1, i, j + 1, used) or self.helper(board, word, index + 1, i, j - 1, used)        used[i][j] = False        return res