Cinte's Leetcode Record
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130. Surrounded Regions

Posted on Edited on In Disqus:
Symbols count in article: 3k Reading time ≈ 3 mins.

130. Surrounded Regions

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class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int m = board.size(), n = board[0].size();
        for(int i = 0; i < n; ++i)
        {
            goNeighbor(board, 0, i);
            goNeighbor(board, m - 1, i);
        }
        for(int i = 0; i < m; ++i)
        {
            goNeighbor(board, i, 0);
            goNeighbor(board, i, n - 1);
        }
        for(int i = 0; i < m; ++i)
            for(int j = 0; j < n; ++j)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '1')
                    board[i][j] = 'O';
            }
    }
private:
    void goNeighbor(vector<vector<char>>& board, int i, int j)
    {
        if(board[i][j] != 'O')
            return;
        board[i][j] = '1';
        if(i > 0)
            goNeighbor(board, i - 1, j);
        if(i < board.size() - 1)
            goNeighbor(board, i + 1, j);
        if(j > 0)
            goNeighbor(board, i, j - 1);
        if(j < board[0].size() - 1)
            goNeighbor(board, i, j + 1);
    }
};

当且仅当和边界相连的O块区域才可能被表留，而在中间的就全都被覆盖，因此先沿着边界检查，把所有这种O区域改成1表示要保留的，然后把里面剩下的O改成X，1改回O就行了

使用迭代 BFS?
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class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int m = board.size(), n = board[0].size();
        queue<pair<int, int>> q;
        for(int i = 0; i < m; ++i)
        {
            for(int j = 0; j < n; ++j)
            {
                if(j != 0 && i != 0 && j != n - 1 && i != m - 1)
                    continue;
                q.push(make_pair(i, j));
                while(!q.empty())
                {
                    auto b = q.front();
                    q.pop();
                    int i = b.first;
                    int j = b.second;
                    if(board[i][j] == 'O')
                    {
                        board[i][j] = '1';
                        if(i > 1)
                            q.push(make_pair(i - 1, j));
                        if(i < m - 2)
                            q.push(make_pair(i + 1, j));
                        if(j > 1)
                            q.push(make_pair(i, j - 1));
                        if(j < n - 2)
                            q.push(make_pair(i, j + 1));
                    }   
                }
            }
        }
        for(int i = 0; i < m; ++i)
            for(int j = 0; j < n; ++j)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                else if(board[i][j] == '1')
                    board[i][j] = 'O';
            }
    }
};

关于stackoverflow https://leetcode.com/problems/surrounded-regions/discuss/41612/A-really-simple-and-readable-C++-solutionuff0conly-cost-12ms/39811

每个递归调用32B存储空间(1个int)（有待考证），然后250*2501623589017459.png

就占用32 * 250 * 250 / 2 = 1000000B = 1M = 栈大小

review

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class Solution {
public:
    void solve(vector<vector<char>>& board) {
        int m = board.size(), n = board[0].size();
        queue<pair<int, int>> q;
        for(int i = 0; i < m; ++i)
        {
            dfsIt(board, m, n, i, 0);
            dfsIt(board, m, n, i, n - 1);
        }
        for(int i = 0; i < n; ++i)
        {
            dfsIt(board, m, n, 0, i);
            dfsIt(board, m, n, m - 1, i);
        }
        for(int i = 0; i < m; ++i)
            for(int j = 0; j < n; ++j)
                if(board[i][j] != '#')
                    board[i][j] = 'X';
                else if(board[i][j] == '#')
                    board[i][j] = 'O';
    }
private:
    inline void dfsIt(vector<vector<char>>& board, int m, int n, int i, int j)
    {
        queue<pair<int, int>> q;
        q.emplace(i, j);
        while(!q.empty())
        {
            int i, j;
            tie(i, j) = q.front();
            q.pop();
            if(i < 0 || i >= m || j < 0 || j >=n)
                continue;
            if(board[i][j] == 'O')
            {
                board[i][j] = '#';
                q.emplace(i + 1, j);
                q.emplace(i - 1, j);
                q.emplace(i, j + 1);
                q.emplace(i, j - 1);
            }
        }
    }
};