Leetcode286墙与门

yeah_panda

于 2020-08-04 12:03:46 发布

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分类专栏： Leetcode 文章标签： python 算法 leetcode

本文链接：https://blog.youkuaiyun.com/weixin_39504758/article/details/107783634

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本文介绍了解决LeetCode上编号为286的题目房间与门的算法实现。该算法使用广度优先搜索策略，从所有门的位置开始，计算每个空房间到最近门的距离，并更新房间矩阵中的距离值。文章详细解释了算法流程，包括初始化搜索队列，进行广度优先搜索，更新距离值等关键步骤。

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286墙与门

#286墙与门
class Solution:
    def wallsAndGates(self, rooms):
        EMPTYROOM = 2**31 - 1
        door = 0
        neighbors = [[-1,0], [1,0], [0,-1], [0,1]]
        if not rooms:
            return
        row = len(rooms)
        col = len(rooms[0])
        searchlist = deque()
        for i in range(row):
            for j in range(col):
                if rooms[i][j] == door:
                    searchlist.append(([i,j], 0))

        while searchlist:
            [n_row, n_col], distance = searchlist.popleft()
            for d_row, d_col in neighbors:
                neighbor_row = d_row + n_row
                neighbor_col = d_col + n_col
                if 0 <= neighbor_row < row and 0<=neighbor_col < col \
                    and rooms[neighbor_row][neighbor_col] == EMPTYROOM:
                    rooms[neighbor_row][neighbor_col] = distance + 1
                    searchlist.append(([neighbor_row,neighbor_col], distance + 1))
        return print(rooms)

data = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1],[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
data1 = []
data2 = [[-1]]
data3 = [[2147483647]]
data4 = [[2147483647,0,2147483647,2147483647,0,2147483647,-1,2147483647]]
solution = Solution()
solution.wallsAndGates(data)