LeetCode 200 岛屿数量 NC109 岛屿数量

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已于 2022-06-23 02:23:11 修改

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分类专栏：数据结构与算法文章标签：算法 leetcode 数据结构

于 2020-12-28 01:21:36 首次发布

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BM57 岛屿数量

给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。

岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。

此外，你可以假设该网格的四条边均被水包围。

示例 1：

输入：grid = [
["1","1","1","1","0"],
["1","1","0","1","0"],
["1","1","0","0","0"],
["0","0","0","0","0"]
]
输出：1

示例 2：

输入：grid = [
["1","1","0","0","0"],
["1","1","0","0","0"],
["0","0","1","0","0"],
["0","0","0","1","1"]
]
输出：3

提示：

m == grid.length
n == grid[i].length
1 <= m, n <= 300
grid[i][j] 的值为 '0' 或 '1'

解题思路：

深度优先搜索

岛屿数量 - 岛屿数量 - 力扣（LeetCode）

Python代码：

class Solution:
    def dfs(self, grid, i, j):
        if not 0 <= i < len(grid) or not 0 <= j < len(grid[0]) or grid[i][j] == '0':
            return None
        grid[i][j] = '0'
        for di, dj in [[0, 1], [0, -1], [1, 0], [-1, 0]]:
            new_i = i + di
            new_j = j + dj
            self.dfs(grid, new_i, new_j)
            
    def numIslands(self, grid: List[List[str]]) -> int:
        m, n = len(grid), len(grid[0])
        res = 0
        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    res += 1
                    self.dfs(grid, i, j)
        return res