LeetCode 200. Number of Islands [Python]

YIZHUO�FENG

于 2019-11-08 14:03:01 发布

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分类专栏： LeetCode 文章标签： Python3 LeetCode

本文链接：https://blog.youkuaiyun.com/yfeng18/article/details/102971875

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本文介绍了一种使用宽度优先搜索算法来计算二维网格中岛屿数量的方法。岛屿由'1'（陆地）组成，被'0'（水域）包围，通过水平或垂直连接形成。文章详细解释了如何遍历整个图，标记访问过的节点，并统计岛屿数量。

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Given a 2d grid map of '1’s (land) and '0’s (water), count the number of islands. An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

宽度优先搜索四个方向遍历整个图
访问过的节点标记为0

class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        count = 0
        if not grid or not grid[0]:
            return count
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    self.bfs(grid, i, j)
                    count += 1
        return count
    
    def bfs(self, grid, i, j):
        queue = collections.deque([(i,j)])
        grid[i][j] = '0'
        while queue:
            x, y = queue.popleft()
            for delta_x, delta_y in [(1,0), (0,1), (-1,0), (0,-1)]:
                new_x, new_y = x + delta_x, y + delta_y
                if not self.valid(grid, new_x, new_y):
                    continue
                queue.append((new_x, new_y))
                grid[new_x][new_y] = '0'
    
    def valid(self, grid, i, j):
        m, n = len(grid), len(grid[0])
        return 0 <= i < m and 0 <= j < n and grid[i][j] == '1'