[leetcode]Surrounded Regions @ Python

原题地址：https://oj.leetcode.com/problems/surrounded-regions/

题意：

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

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

For 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

解题思路：这道题可以使用BFS和DFS两种方法来解决。DFS会超时。BFS可以AC。从边上开始搜索，如果是'O'，那么搜索'O'周围的元素，并将'O'置换为'D'，这样每条边都DFS或者BFS一遍。而内部的'O'是不会改变的。这样下来，没有被围住的'O'全都被置换成了'D'，被围住的'O'还是'O'，没有改变。然后遍历一遍，将'O'置换为'X'，将'D'置换为'O'。

dfs代码，因为递归深度的问题会爆栈：

class Solution:
    # @param board, a 9x9 2D array
    # Capture all regions by modifying the input board in-place.
    # Do not return any value.
    def solve(self, board):
        def dfs(x, y):
            if x< or x>m- or y< or y>n- or board[x][y]!='O':return
            board[x][y] = 'D'
            dfs(x-, y)
            dfs(x+, y)
            dfs(x, y+)
            dfs(x, y-)

        if len(board) == : return
        m = len(board); n = len(board[])
        for i in range(m):
            dfs(i, ); dfs(i, n-)
        for j in range(, n-):
            dfs(, j); dfs(m-, j)
        for i in range(m):
            for j in range(n):
                if board[i][j] == 'O': board[i][j] == 'X'
                elif board[i][j] == 'D': board[i][j] == 'O'

bfs代码：

class Solution:
    # @param board, a 9x9 2D array
    # Capture all regions by modifying the input board in-place.
    # Do not return any value.
    def solve(self, board):
        def fill(x, y):
            if x< or x>m- or y< or y>n- or board[x][y] != 'O': return
            queue.append((x,y))
            board[x][y]='D'
        def bfs(x, y):
            if board[x][y]=='O':queue.append((x,y)); fill(x,y)
            while queue:
                curr=queue.pop(); i=curr[]; j=curr[]
                fill(i+,j);fill(i-,j);fill(i,j+);fill(i,j-)
        if len(board)==: return
        m=len(board); n=len(board[]); queue=[]
        for i in range(n):
            bfs(,i); bfs(m-,i)
        for j in range(, m-):
            bfs(j,); bfs(j,n-)
        for i in range(m):
            for j in range(n):
                if board[i][j] == 'D': board[i][j] = 'O'
                elif board[i][j] == 'O': board[i][j] = 'X'