題目

130. 被圍繞的區域

我的思路

找到四周邊界上是‘O’的位置，然後向內部深搜或者廣搜。把連通的‘O’修改為‘K’。之後再把所有‘K’置‘O’，而未被修改過的‘O’也就是被包圍的不連通的，置‘X’。

我的實現

class Solution {
public:
    const int dx[4] = {1, -1, 0, 0};
    const int dy[4] = {0, 0, 1, -1};

    void solve(vector<vector<char
 
>>& board) {
        int n = board.size();
        if (n == 0) {
            return;
        }
        int m = board[0].size();
        queue<pair<int, int>> que;
        for (int i = 0; i < n; i++) {
            if (board[i][0] == 'O') {
                que.emplace(i, 0);
            }
            
 
if (board[i][m - 1] == 'O') {
                que.emplace(i, m - 1);
            }
        }
        for (int i = 1; i < m - 1; i++) {
            if (board[0][i] == 'O') {
                que.emplace(0, i);   
            }
            if (board[n - 1][i] == 'O') {
                que.emplace(n - 1
 
, i);
            }
        }
        while (!que.empty()) {
            int x = que.front().first, y = que.front().second;
            que.pop();
            board[x][y] = 'A';
            for (int i = 0; i < 4; i++) {
                int mx = x + dx[i], my = y + dy[i];
                if (mx < 0 || my < 0 || mx >= n || my >= m || board[mx][my] != 'O') {
                    continue;
                }
                que.emplace(mx, my);
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (board[i][j] == 'A') {
                    board[i][j] = 'O';
                } else if (board[i][j] == 'O') {
                    board[i][j] = 'X';
                }
            }
        }
    }
};

深搜or廣搜

時間複雜度是on，矩陣的大小

拓展學習