Number_of_Islands.md

200. 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

Method:

dfs stack, (bfs deque similar)

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if not grid:
            return 0
        
        m, n=len(grid), len(grid[0])
        stack=[]
        d={}
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    d[(i, j)]=i*n+j
                    stack.append((i, j))
                    
        while stack:
            i, j=stack.pop()
            for (di, dj) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                if i+di>=0 and i+di<m and j+dj>=0 and j+dj<n and grid[i+di][j+dj]=='1':
                    d[(i+di, j+dj)]=d[(i, j)]
                    grid[i+di][j+dj]='2'
                    stack.append((i+di, j+dj))
                                        
        return len(set(d.values()))

Solution:

一次遍历即可：

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if not grid:
            return 0
        
        m, n=len(grid), len(grid[0])
        stack=[]
        count=0
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    stack.append((i, j))
                    count+=1
                    while stack:
                        x, y=stack.pop()
                        for (dx, dy) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                            if x+dx>=0 and x+dx<m and y+dy>=0 and y+dy<n and grid[x+dx][y+dy]=='1':
                                grid[x+dx][y+dy]='0'
                                stack.append((x+dx, y+dy))                                        
        return count

recursion:

class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if not grid:
            return 0
        
        def dfs(i, j):
            for (di, dj) in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                if i+di>=0 and i+di<m and j+dj>=0 and j+dj<n and grid[i+di][j+dj]=='1':
                    grid[i+di][j+dj]='0'
                    dfs(i+di, j+dj)
                    
        m, n=len(grid), len(grid[0])
        count=0
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    count+=1
                    dfs(i, j)                         
        return count

union find:

class UF:
    def __init__(self, val):
        self.parent=[0]*val
        self.cnt=0
        
    def find(self, node):
        if self.parent[node]!=node:
            self.parent[node]=self.find(self.parent[node])
        return self.parent[node]
    
    def union(self, p, q):
        r1, r2=self.find(p), self.find(q)
        if r1!=r2:
            self.parent[r1]=r2
            self.cnt-=1


class Solution(object):
    def numIslands(self, grid):
        """
        :type grid: List[List[str]]
        :rtype: int
        """
        if not grid:
            return 0
                    
        m, n=len(grid), len(grid[0])
        uf=UF(m*n)
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    uf.parent[i*n+j]=i*n+j
                    uf.cnt+=1
                    
        for i in range(m):
            for j in range(n):
                if grid[i][j]=='1':
                    for di, dj in [[0,-1],[0,1],[1,0],[-1,0]]:
                        if 0<=i+di<m and 0<=j+dj<n and grid[i+di][j+dj]=='1':
                            uf.union(i*n+j, (i+di)*n+(j+dj))
        return uf.cnt