【python/leetcode/130/M】Surrounded Regions

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题目

https://leetcode.com/problems/surrounded-regions/

基本思路

转换一下思路，找出哪些O是没有被X包围的。在面板四周的O肯定是没有被X包围的，与它们相连的O也是没有被包围的，其它的O都是被X包围的。

问题简化为将与四周的O相连的O都找出来，这些点不用变，其它点都变为X。

首先将四周的O压入栈内，依次访问栈内元素，并将它们标记，接着去判断它们四周的元素是否也是O，如果是且没有被标记过，则将其压入栈中。当遍历完栈中的元素后，将有标记的元素变为O，其余都是X。

实现代码

class Solution(object):
    def solve(self, board):
        """
        :type board: List[List[str]]
        :rtype: void Do not return anything, modify board in-place instead.
        """
        if not board or not board[0]:
            return 
        n,m = len(board),len(board[0])
        
        # 找出所有的边界O，将其押入栈中
        stack = []
        for i in range(n):
            for j in range(m):
                if ((i in (0, n - 1)) or (j in (0, m - 1))) and board[i][j] == 'O':
                    stack.append((i, j))
        
        # 标记所有能联系到边界的O的O
        while stack:
            r,c = stack.pop(0)
            
            if 0<=r<n and 0<=c<m and board[r][c] == 'O':
                board[r][c] = 'M'
                if r-1>=0 and board[r-1][c] == 'O':
                    stack.append((r-1,c))
                if r+1<n and board[r+1][c] == 'O':
                    stack.append((r+1,c))
                if c-1>=0 and board[r][c-1] == 'O':
                    stack.append((r,c-1))
                if c+1<m and board[r][c+1] == 'O':
                    stack.append((r,c+1))
        
        # 更新
        for i in range(n):
            for j in range(m):
                if board[i][j] == 'M':
                    board[i][j] = 'O'
                else:
                    board[i][j] = 'X'