0200：岛屿数量（★）

2015-07-19 约 314 字预计阅读 1 分钟次阅读

力扣第 200 题

题目

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'

分析

可以用 dfs/bfs 遍历找到每一个岛
连通问题更通用的办法是用并查集
- 将相邻的陆地连通
- 最终陆地连通块的个数即是所求

解答

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class Solution:
    def numIslands(self, grid: List[List[str]]) -> int:
        def find(x):
            if f[x] != x:
                f[x] = find(f[x])
            return f[x]

        def union(x, y):
            f[find(x)] = find(y)

        m, n = len(grid), len(grid[0])
        f = list(range(m*n))
        for i, j in product(range(m), range(n)):
            if grid[i][j] == '1':
                if i and grid[i-1][j] == '1':
                    union(i*n+j,(i-1)*n+j)
                if j and grid[i][j-1] == '1':
                    union(i*n+j,i*n+j-1)
        return sum(find(x)==x for x in range(m*n) if grid[x//n][x%n]=='1')

244 ms

目录

0200：岛屿数量（★）

题目

分析

解答