Number of Islands

最新推荐文章于 2025-05-07 17:27:01 发布

原创最新推荐文章于 2025-05-07 17:27:01 发布 · 280 阅读

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本文介绍了一种计算二维网格中岛屿数量的方法。使用并查集（Union-Find）算法和深度优先搜索（DFS）两种方式实现。并查集算法通过连接相邻陆地来减少岛屿数量；而DFS则将每个岛屿的所有部分标记为已访问。

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.

Example 1:
11110
11010
11000
00000
Answer: 1

Example 2:
11000
11000
00100
00011
Answer: 3

打算用union-find的方法，用dfs迭代时有可能内存溢出。

class UF(object):
    def union(self, a, b, tmp_grid, sz):
        p = self.root(a, tmp_grid)
        q = self.root(b, tmp_grid)
        if p == q:
            return
        if sz[p] > sz[q]:
            tmp_grid[q] = p
            sz[p] += sz[q]
        else:
            tmp_grid[p] = q
            sz[q] += sz[p]

    def root(self, a, tmp_grid):
        while(tmp_grid[a] != a):
            a = tmp_grid[a]

        return a


class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if len(grid) == 0:
            return 0
        tmp_grid = [i for i in range(len(grid) * len(grid[0]))]
        uf = UF()
        m = len(grid)
        n = len(grid[0])
        sz = [1] * (m*n)

        for i in range(m):
            for j in range(n):
                if grid[i][j] == '1':
                    if j+1 < n:
                        if grid[i][j+1] == '1':
                            uf.union(i*n+j, i*n+j+1, tmp_grid, sz)
                    if i+1 < m:
                        if grid[i+1][j] == '1':
                            uf.union(i*n+j, (i+1)*n+j, tmp_grid, sz)

        count = 0
        for i in range(len(tmp_grid)):
            if tmp_grid[i] == i and grid[i/n][i%n] == '1':
                count += 1

        return count

在网上看到比较好的解法是用dfs。如果当前grid为1，则将其设置为2，并将其周围所有的1都设置为2。这样在搜索过程中遇到的所有的1就是连通量的数目。

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if len(grid) == 0:
            return 0

        count = 0    
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '1':
                    count += 1
                    self.dfs(i, j, grid)

        return count
    def dfs(self, i, j, grid):
        if i < 0 or i >= len(grid) or j < 0 or j >= len(grid[0]):
            return
        if grid[i][j] == '1':
            grid[i][j] = '2'
            self.dfs(i, j+1, grid)
            self.dfs(i, j-1, grid)
            self.dfs(i-1, j, grid)
            self.dfs(i+1, j, grid)