200. 岛屿数量&&130. 被围绕的区域

给定一个由 '1'（陆地）和 '0'（水）组成的的二维网格，计算岛屿的数量。一个岛被水包围，并且它是通过水平方向或垂直方向上相邻的陆地连接而成的。你可以假设网格的四个边均被水包围。

示例 1:

输入:
11110
11010
11000
00000

输出: 1
示例 2:

输入:
11000
11000
00100
00011

输出: 3

1.dfs。遇到一个1就其上下左右4个方向进行搜索，将遍历过的1置0。最后的结果就是dfs启动的次数

2.bfs

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

class Solution:
    def numIslands(self, grid: [[str]]) -> int:
        def bfs(grid, i, j):
            queue = [[i, j]]
            while queue:
                [i, j] = queue.pop(0)
                if 0 <= i < len(grid) and 0 <= j < len(grid[0]) and grid[i][j] == '1':
                    grid[i][j] = '0'
                    queue += [[i + 1, j], [i - 1, j], [i, j - 1], [i, j + 1]]
        count = 0
        for i in range(len(grid)):
            for j in range(len(grid[0])):
                if grid[i][j] == '0': continue
                bfs(grid, i, j)
                count += 1
        return count

给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。

找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

示例:

X X X X
X O O X
X X O X
X O X X
运行你的函数后，矩阵变为：

X X X X
X X X X
X X X X
X O X X

1.和上一题一个思路，从边界开始搜索，搜索过的地方做一个标记

class Solution(object):
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: None Do not return anything, modify board in-place instead.
        """
        flag = [[True for _ in range(len(board[0]))] for _ in range(len(board))]
        def dfs(i,j):
            print(i,j)
            if not 0<=i<len(board) or not 0<=j<len(board[0]) or board[i][j]=='X' or not flag[i][j]:
                return 
            flag[i][j]=False
            dfs(i-1,j)
            dfs(i+1,j)
            dfs(i,j-1)
            dfs(i,j+1)
        for i in range(len(board)):
            for j in range(len(board[0])):
                if (i==0 or (i==len(board)-1) or j==0 or (j==len(board[0])-1)) and board[i][j]=='O':
                    dfs(i,j)
        for i in range(len(board)):
            for j in range(len(board[0])):
                if flag[i][j]:
                    board[i][j]='X'