【Leetcode 回溯法、深搜】单词搜索（79）

题目

给定一个二维网格和一个单词，找出该单词是否存在于网格中。

单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。

示例:

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

给定 word = "ABCCED", 返回 true
给定 word = "SEE", 返回 true
给定 word = "ABCB", 返回 false

提示：

• board 和 word 中只包含大写和小写英文字母。
• 1 <= board.length <= 200
• 1 <= board[i].length <= 200
• 1 <= word.length <= 10^3

解答

# encoding: utf-8


# 典型的深搜题目
# Time: O(m·n), Space: O(m·n)
class Solution:
    def exist(self, board, word):
        def dfs(x, y, s_index):
            if s_index == len(word):  # 匹配完成了
                return True
            for i in range(4):
                tx = x + next[i][0]
                ty = y + next[i][1]
                if 0 <= tx < rows and 0 <= ty < columns and board[tx][ty] == word[s_index]:
                    temp = board[tx][ty]
                    board[tx][ty] = '/'
                    if dfs(tx, ty, s_index + 1):
                        return True
                    board[tx][ty] = temp
            return False

        if not board:
            return False
        next = [  # 右下左上
            [0, 1],
            [1, 0],
            [0, -1],
            [-1, 0]
        ]
        columns, rows = len(board[0]), len(board)
        # book = [[0 for _ in range(columns)] for _ in range(rows)]  # 标记

        s_index = 0
        for i in range(rows):
            for j in range(columns):
                if board[i][j] == word[s_index]:
                    temp = board[i][j]  # 不用book标记了，存入临时变量并修改board[i][j]为特殊字符，和原来自己的book标记思路相比，节省了O(mn)的空间
                    board[i][j] = '/'
                    if dfs(i, j, s_index + 1):
                        return True
                    board[i][j] = temp
        return False


s = Solution()
ans = s.exist([["a", "b", "c", "e"],
               ["s", "f", "c", "s"],
               ["a", "d", "e", "e"]], 'bfce')
print(ans)  # True