LeetCode解题之Surrounded Regions

原题

有一个二维的面板，上面由”X”或者”O”填充。现在要求将被”X”包围的”O”都改成”X”。

注意点：

• 无

例子:

输入:

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

输出:

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

解题思路

直接去找被X包围的O比较麻烦，不如转换一下思路，找出哪些O是没有被X包围的。首先在面板四周的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。

AC源码

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 = len(board)
        m = len(board[0])
        queue = []
        # Get all ‘O‘ on edge
        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‘:
                    queue.append((i, j))
        # Mark all ‘O‘ which can connect to ‘O‘ on edge
        while queue:
            r, c = queue.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‘:
                    queue.append((r - 1, c))
                if r + 1 < n and board[r + 1][c] == ‘O‘:
                    queue.append((r + 1, c))
                if c - 1 >= 0 and board[r][c - 1] == ‘O‘:
                    queue.append((r, c - 1))
                if c + 1 < m and board[r][c + 1] == ‘O‘:
                    queue.append((r, c + 1))
        # Update characters
        for i in range(n):
            for j in range(m):
                if board[i][j] == ‘M‘:
                    board[i][j] = ‘O‘
                else:
                    board[i][j] = ‘X‘


if __name__ == "__main__":
    board = [
        [‘X‘, ‘X‘, ‘X‘, ‘X‘],
        [‘X‘, ‘O‘, ‘O‘, ‘X‘],
        [‘X‘, ‘X‘, ‘O‘, ‘X‘],
        [‘X‘, ‘O‘, ‘X‘, ‘X‘]
    ]
    expected_board = [
        [‘X‘, ‘X‘, ‘X‘, ‘X‘],
        [‘X‘, ‘X‘, ‘X‘, ‘X‘],
        [‘X‘, ‘X‘, ‘X‘, ‘X‘],
        [‘X‘, ‘O‘, ‘X‘, ‘X‘]
    ]
    Solution().solve(board)
    assert board == expected_board

欢迎查看我的Github (https://github.com/gavinfish/LeetCode-Python) 来获得相关源码。