力扣-79 单词搜索

最新推荐文章于 2025-06-29 10:48:52 发布

原创最新推荐文章于 2025-06-29 10:48:52 发布 · 124 阅读

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该博客介绍了一个Python类`Solution`，实现了回溯算法来检查给定的二维字符矩阵中是否存在指定的单词。算法通过遍历矩阵并标记已访问的位置，从匹配到的第一个字符开始进行深度优先搜索，探索所有可能的路径。当找到完整的单词时，返回`True`，否则返回`False`。

# 使用回溯算法
class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        # 有个二维列表记录当前位置是否走过
        # 记录走过的路径，如果word在走过的路径中，则返回true
        self.res = False
        # m * n 的矩阵
        m = len(board)
        n = len(board[0])
        mask = self.create_mask(m,n)
        # mask[0][0]=1
        for i in range(m):
            for j in range(n):
                # 先找到word的第一个字符，然后开始回溯
                if board[i][j] == word[0]:
                    cur = board[i][j]
                    mask[i][j] = 1
                    self.drawback(board[:], i, j, cur[:], mask[:], word[:])
                    mask[i][j] = 0
        return self.res

    def drawback(self, board, i, j, cur, mask, word):
        if cur != word[:len(cur)]:
            return
        if cur == word:
            if cur == word:
                self.res = True
            return
        # 上、下、左、右四个方位
        # 左
        if j-1 >= 0 :
            if mask[i][j-1] == 0:
                cur += board[i][j-1]
                mask[i][j-1] = 1
                self.drawback(board[:], i, j-1, cur[:], mask[:], word[:])
                cur = cur[:-1]
                mask[i][j-1] = 0
        # 右
        if j+1 < len(board[0]):
            if mask[i][j+1] == 0:
                cur += board[i][j+1]
                mask[i][j+1] = 1
                self.drawback(board[:], i, j+1, cur[:], mask[:], word[:])
                cur = cur[:-1]
                mask[i][j+1] = 0

        # 上
        if i-1 >= 0:
            if mask[i-1][j] == 0:
                cur += board[i-1][j]
                mask[i-1][j] = 1
                self.drawback(board[:], i-1, j, cur[:], mask[:], word[:])
                cur = cur[:-1]
                mask[i-1][j] = 0

        # 下
        if i+1 < len(board):
            if mask[i+1][j] == 0:
                cur += board[i+1][j]
                mask[i+1][j] = 1
                self.drawback(board[:], i+1, j, cur[:], mask[:], word[:])
                cur = cur[:-1]
                mask[i+1][j] = 0
        return

    
    def create_mask(self,m,n):
        mask =  [[0 for col in range(n)] for row in range(m)]
        return mask[:]