"""
79. Word Search
MediumShare
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.Example:board =
[['A','B','C','E'],['S','F','C','S'],['A','D','E','E']
]Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.
"""

class Solution(object):def searchWordinBoard(self, board, word, len_word, n, i, j, pathset):""":type board:list[list[str]]:type word:str:type n,i,j:int:type pathset:set:rtype:bool几个参数分别代表矩阵,单词,单词长度,当前字母,当前矩阵位置,已经经过的路径集合"""if i<0 or j<0 or i>=len(board) or j>=len(board[0]):return Falseif word[n]!=board[i][j] or (i,j) in pathset:return Falseif n == len_word-1:return Truepathset.add((i,j))result = self.searchWordinBoard(board, word, len_word, n+1, i-1, j, pathset) or \self.searchWordinBoard(board, word, len_word, n+1, i, j-1, pathset) or \self.searchWordinBoard(board, word, len_word, n+1, i+1, j, pathset) or \self.searchWordinBoard(board, word, len_word, n+1, i, j+1, pathset)pathset.remove((i,j))return resultdef exist(self, board, word):""":type board: List[List[str]]:type word: str:rtype: bool"""if word == "":return Trueif not board or not board[0]:return Falselen_row, len_col = len(board), len(board[0])for i in range(len_row):for j in range(len_col):if word[0] != board[i][j]:continueif self.searchWordinBoard(board, word, len(word), 0, i, j, set()):return Truereturn False