Number of Islands

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 aclass 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)