[LeetCode] 130. Surrounded Regions_Medium tag: DFS/BFS

Given a 2D board containing 'X' and 'O' (the letter O), capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

Example:

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

After running your function, the board should be:

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

Explanation:

Surrounded regions shouldn’t be on the border, which means that any 'O' on the border of the board are not flipped to 'X'. Any 'O' that is not on the border and it is not connected to an 'O' on the border will be flipped to 'X'. Two cells are connected if they are adjacent cells connected horizontally or vertically.

 

这个题目思路实际上跟[LeetCode] 733. Flood Fill_Easy tag: BFS很像, 只不过我们将array遍历两边,第一次遍历边框, 如果是'O' 将所有相邻的'O' 都标记为visited, 然后第二次遍历

如果没有标记为visited, 将其换为'X'即可.

 

1. Constriants

1) None or n == 0

2) element will be only 'X' or 'O'

 

2. Ideas

DFS/BFS     T: O(m*n)    S; O(m*n)

 

3. Code

3.1) DFS

class Solution:
    def surroundRegion(self, board):
        if not board or len(board[0]) == 0: return
        lr, lc , visited = len(board), len(board[0]), set()
        def dfs(r, c):
            if 0 <= r < len(board) and 0 <= c < len(board[0]):
                if (r,c) not in visited and board[r][c] == 'O':
                    visited.add((r,c))
                    dfs(r+1, c)
                    dfs(r-1, c)
                    dfs(r,c+1)
                    dfs(r,c-1)
        
        for i in range(lr):
            for j in range(lc):
                if (i== 0 or i == lr-1 or j == 0 or j == lc -1 ) and board[i][j] == 'O' and (i,j) not in visted:
                    dfs(i,j)
        for i in range(lr):
            for j in range(lc):
                if board[i][j] == 'O' and (i,j) not in visited:
                    board[i][j] = 'X'

 

3.2) BFS

class Solution:
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        if not board or len(board[0]) == 0 : return 
        lr, lc, queue, visited = len(board), len(board[0]), collections.deque(), set()
        for i in range(lr):
            for j in range(lc):
                if (i == 0 or i == lr - 1  or j == 0 or j == lc -1) and board[i][j] == 'O' :
                    queue.append((i,j))
                    visited.add((i,j))
        dirs = [(0,1), (0,-1), (1,0), (-1,0)]
        while queue:
            pr, pc = queue.popleft()
            for c1, c2 in dirs:
                nr, nc = pr + c1 , pc + c2
                if 0 <= nr < lr and 0 <= nc <lc and (nr, nc) not in visited and board[nr][nc] == 'O':
                    queue.append((nr, nc))
                    visited.add((nr, nc))
        
        for i in range(lr):
            for j in range(lc):
                if board[i][j] == 'O' and (i,j) not in visited: # first check (i, j) not in visited will speed up the process a lot. 
                    board[i][j] = 'X'