LeetCode Surrounded Regions
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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 XX O O XX X O XX O X X
输出:
X X X XX X X XX X X XX 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。
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