leetcode-word search

 Word search

Given a two-dimensional grid and a word, find out whether the word exists in the grid.

Words must be in alphabetical order and formed by letters in adjacent cells, where "adjacent" cells are those that are adjacent horizontally or vertically. Letters in the same cell are not allowed to be reused.

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

prompt:

Only uppercase and lowercase English letters are contained in board and word.
1 <= board.length <= 200
1 <= board[i].length <= 200
1 <= word.length <= 10^3
Passed 108,459 Submitted 249,098

 

Code

class Solution:
    def equal_c(self, board, mark, c, row, col):
        if row >= len(board) or row < 0:
            return False
        if col >= len(board[row]) or col < 0:
            return False
        if mark[row][col] == 0 and board[row][col]==c:
            return True
        return False

    def exist(self, board, word):
        word_clist = list(word)
        for row in range(len(board)):
            for col in range(len(board[row])):
                if word_clist[0]!=board[row][col]:
                    continue
                if len(word_clist)==1:
                    return True
                mark = [[0] * len(item) for item in board] # 用来标记是否已经选中
                choose_list = [''] * len(word_clist)
                cur_word_i = 0
                choose_list[0] = (row, col)
                mark[row][col] = 1
                cur_word_i += 1
                temp_store = [[] for i in range(len(word_clist))] # 暂存每一位未遍历的点
                temp_store[cur_word_i].extend([(row, col+1),(row+1, col),(row, col-1),(row-1, col)])
                while cur_word_i > 0:
                    if len(temp_store[cur_word_i])==0:
                        cur_word_i -= 1
                        pre_pos = choose_list[cur_word_i]
                        mark[pre_pos[0]][pre_pos[1]] = 0
                        choose_list[cur_word_i] = ''
                        continue
                    pos = temp_store[cur_word_i][0]
                    temp_store[cur_word_i].remove(pos)
                    if self.equal_c(board, mark, word_clist[cur_word_i], pos[0], pos[1]): #匹配上，进行下一位置搜索
                        if cur_word_i == len(word_clist) - 1:
                            return True
                        choose_list[cur_word_i] = pos
                        mark[pos[0]][pos[1]] = 1
                        cur_word_i += 1
                        temp_store[cur_word_i].extend([(pos[0], pos[1]+1),(pos[0]+1, pos[1]),(pos[0], pos[1]-1),
                                                       (pos[0]-1, pos[1])])
        return False