Leetcode200岛屿数量

最新推荐文章于 2025-06-02 14:28:44 发布

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最新推荐文章于 2025-06-02 14:28:44 发布

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分类专栏： Leetcode 文章标签： dfs bfs leetcode 队列 queue

本文链接：https://blog.youkuaiyun.com/weixin_39504758/article/details/107783683

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本文深入解析了LeetCode上的经典题目“岛屿数量”(LeetCode200)，通过深度优先搜索（DFS）和广度优先搜索（BFS）两种算法详细阐述了解决方案。文章首先介绍了问题背景，随后提供了Python实现代码，并对比分析了两种算法的优缺点。

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Leetcode200岛屿数量

bfs

class Solution:
    def numIslands(self, grid):
        if not grid or not grid[0]:
            return 0
        self.m, self.n = len(grid), len(grid[0])
        islands = 0
        for row in range(self.m):
            for col in range(self.n):
                if grid[row][col] == "1":
                    islands += 1
                    grid[row][col] = '0'
                    self.bfs(grid, row, col)

        return islands

    def bfs(self, grid, row, col):
        queue = collections.deque([(row, col)])
        while queue:
            row, col = queue.popleft()
            for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:
                if 0 <= x < self.m and 0 <= y < self.n and grid[x][y] == "1":
                    queue.append((x, y))
                    grid[x][y] = '0'

dfs

class Solution:
    def numIslands(self, grid):
        if not grid or not grid[0]:
            return 0
        self.m, self.n = len(grid), len(grid[0])
        islands = 0
        for row in range(self.m):
            for col in range(self.n):
                if grid[row][col] == "1":
                    islands += 1
                    self.dfs(grid, row, col)

        return islands

    def dfs(self, grid, row, col):
        grid[row][col] = "0"
        for x, y in [(row - 1, col), (row + 1, col), (row, col - 1), (row, col + 1)]:
            if 0 <= x < self.m and 0 <= y < self.n and grid[x][y] == "1":
                self.dfs(grid, x, y)