[LeetCode] 130. 被围绕的区域

题目链接 : https://leetcode-cn.com/problems/surrounded-regions/

题目描述:

给定一个二维的矩阵，包含 'X' 和 'O'（字母 O）。

找到所有被 'X' 围绕的区域，并将这些区域里所有的 'O' 用 'X' 填充。

示例:

示例:

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

运行你的函数后，矩阵变为：

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

解释:

被围绕的区间不会存在于边界上，换句话说，任何边界上的 'O' 都不会被填充为 'X'。 任何不在边界上，或不与边界上的 'O' 相连的 'O' 最终都会被填充为 'X'。如果两个元素在水平或垂直方向相邻，则称它们是“相连”的。

思路:

从边界出发把,先把边界上和O连通点找到, 把这些变成B,然后遍历整个board把O变成X, 把B变成O

如下图所示

所以这样就有2种方法

思路一: DFS

思路二: BFS

还有一种就是, 把边界O并且与它连通这些点分在一起

思路三 : 并查集

代码:

思路一:

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        if not board or not board[0]:
            return
        row = len(board)
        col = len(board[0])
        def dfs(i, j):
            board[i][j] = "B"
            for x, y in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                tmp_i = i + x
                tmp_j = j + y
                if 1 <= tmp_i < row and 1 <= tmp_j < col and board[tmp_i][tmp_j] == "O":
                    dfs(tmp_i, tmp_j)
        for j in range(col):
            # 第一行
            if board[0][j] == "O":
                dfs(0, j)
            # 最后一行
            if board[row - 1][j] == "O":
                dfs(row - 1, j)
        for i in range(row):
            # 第一列
            if board[i][0] == "O":
                dfs(i, 0)
            # 最后一列
            if board[i][col-1] == "O":
                dfs(i, col - 1)
        for i in range(row):
            for j in range(col):
                # O 变成 X
                if board[i][j] == "O":
                    board[i][j] = "X"
                # B 变成 O
                if board[i][j] == "B":
                    board[i][j] = "O"

java

class Solution {
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0] == null || board[0].length == 0) return;
        int row = board.length;
        int col = board[0].length;
        for (int j = 0; j < col; j++) {
            // 第一行
            if (board[0][j] == 'O') dfs(0, j, board, row, col);
            // 最后一行
            if (board[row - 1][j] == 'O') dfs(row - 1, j, board, row, col);
        }
        for (int i = 0; i < row; i++) {
            // 第一列
            if (board[i][0] == 'O') dfs(i, 0, board, row, col);
            // 最后一列
            if (board[i][col - 1] == 'O') dfs(i, col - 1, board, row, col);
        }
        // 转变
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == 'B') board[i][j] = 'O';
            }
        }
    }
    private void dfs(int i, int j, char[][] board, int row, int col) {
        board[i][j] = 'B';
        for (int[] dir : dirs) {
            int tmp_i = dir[0] + i;
            int tmp_j = dir[1] + j;
            if (tmp_i < 0 || tmp_i >= row || tmp_j < 0 || tmp_j >= col || board[tmp_i][tmp_j] != 'O') continue;
            dfs(tmp_i, tmp_j, board, row, col);
        }
    }
}

思路二:

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        if not board or not board[0]:
            return
        row = len(board)
        col = len(board[0])
        def bfs(i, j):
            from collections import deque
            queue = deque()
            queue.appendleft((i, j))
            while queue:
                i, j = queue.pop()
                if 0 <= i < row and 0 <= j < col and board[i][j] == "O":
                    board[i][j] = "B"
                    for x, y in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                        queue.appendleft((i + x, j + y))
        for j in range(col):
            # 第一行
            if board[0][j] == "O":
                bfs(0, j)
            # 最后一行
            if board[row - 1][j] == "O":
                bfs(row - 1, j)
        for i in range(row):
            if board[i][0] == "O":
                bfs(i, 0)
            if board[i][col - 1] == "O":
                bfs(i, col - 1)
        for i in range(row):
            for j in range(col):
                if board[i][j] == "O":
                    board[i][j] = "X"
                if board[i][j] == "B":
                    board[i][j] = "O"

java

class Solution {
    int[][] dirs = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    private static class Point {
        int x, y;
        Point(int x, int y) {
            this.x = x;
            this.y = y;
        }
    }
    public void solve(char[][] board) {
        if (board == null || board.length == 0 || board[0] == null || board[0].length == 0) return;
        int row = board.length;
        int col = board[0].length;
        for (int j = 0; j < col; j++) {
            // 第一行
            if (board[0][j] == 'O') bfs(0, j, board, row, col);
            // 最后一行
            if (board[row - 1][j] == 'O') bfs(row - 1, j, board, row, col);
        }
        for (int i = 0; i < row; i++) {
            // 第一列
            if (board[i][0] == 'O') bfs(i, 0, board, row, col);
            // 最后一列
            if (board[i][col - 1] == 'O') bfs(i, col - 1, board, row, col);
        }
        // 转变
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == 'B') board[i][j] = 'O';
            }
        }
    }
    private void bfs(int i, int j, char[][] board, int row, int col) {
        Deque<Point> queue = new LinkedList<>();
        queue.offer(new Point(i, j));
        while (!queue.isEmpty()) {
            Point tmp = queue.poll();
            if (tmp.x >= 0 && tmp.x < row && tmp.y >= 0 && tmp.y < col && board[tmp.x][tmp.y] == 'O') {
                board[tmp.x][tmp.y] = 'B';
                for (int[] dir : dirs) queue.offer(new Point(tmp.x + dir[0], tmp.y + dir[1]));
            }
        }
    }
}

思路三:

class Solution:
    def solve(self, board: List[List[str]]) -> None:
        """
        Do not return anything, modify board in-place instead.
        """
        f = {}
        def find(x):
            f.setdefault(x, x)
            if f[x] != x:
                f[x] = find(f[x])
            return f[x]
        def union(x, y):
            f[find(y)] = find(x)
        if not board or not board[0]:
            return
        row = len(board)
        col = len(board[0])
        dummy = row * col
        for i in range(row):
            for j in range(col):
                if board[i][j] == "O":
                    if i == 0 or i == row - 1 or j == 0 or j == col - 1:
                        union(i * col + j, dummy)
                    else:
                        for x, y in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
                            if board[i + x][j + y] == "O":
                                union(i * col + j, (i + x) * col + (j + y))
        for i in range(row):
            for j in range(col):
                if find(dummy) == find(i * col + j):
                    board[i][j] = "O"
                else:
                    board[i][j] = "X"