[LintCode] Surrounded Regions 包围区域

最新推荐文章于 2020-05-03 05:58:04 发布

转载最新推荐文章于 2020-05-03 05:58:04 发布 · 97 阅读

本文介绍了解决LeetCode上围困区域问题的两种方法。该问题要求在一个包含'X'和'O'的二维棋盘中，将所有被'X'包围的'O'区域翻转为'X'。通过深度优先搜索(DFS)策略标记边界上的'O'及其相连的'O'，避免这些区域被错误地翻转。

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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.
Example

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

After capture all regions surrounded by 'X', the board should be:

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

LeetCode上的原题，请参见我之前的博客Surrounded Regions。

解法一：

class Solution {
public:
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    void surroundedRegions(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return;
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>> &board, int x, int y) {
        int m = board.size(), n = board[0].size();
        vector<vector<int>> dir{{0,-1},{-1,0},{0,1},{1,0}};
        board[x][y] = '$';
        for (int i = 0; i < dir.size(); ++i) {
            int dx = x + dir[i][0], dy = y + dir[i][1];
            if (dx >= 0 && dx < m && dy >= 0 && dy < n && board[dx][dy] == 'O') {
                dfs(board, dx, dy);
            }
        }
    }
};

解法二：

class Solution {
public:
    /**
     * @param board a 2D board containing 'X' and 'O'
     * @return void
     */
    void surroundedRegions(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return;
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (board[i][j] == 'O') dfs(board, i , j);
                }
            }
        }
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == 'O') board[i][j] = 'X';
                if (board[i][j] == '$') board[i][j] = 'O';
            }
        }
    }
    void dfs(vector<vector<char>> &board, int i, int j) {
        int m = board.size(), n = board[0].size();
        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != 'O') return;
        board[i][j] = '$';
        dfs(board, i + 1, j);
        dfs(board, i - 1, j);
        dfs(board, i, j + 1);
        dfs(board, i, j - 1);
    }
};