LeetCode 417. 太平洋大西洋水流问题

方法一：深度优先搜索

思路：由海洋向内dfs，如果某个点被两边海洋都dfs（向海拔高的地方dfs）到了，则该点可以流向两个海洋。

class Solution {
     
private:
	int dx[4] = {
      1,0,-1,0 }, dy[4] = {
      0,1,0,-1 }, m, n;
public:
	vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
     
		vector<vector<int>> ans;
		if (matrix.empty() || matrix[0].empty()) return ans;//特判
		m = matrix.size(), n = matrix[0].size();
		vector<vector<bool>> pacific(m, vector<bool>(n, false));
		vector<vector<bool>> atlantic(m, vector<bool>(n, false));
		for (int i = 0; i < m; ++i) {
     //列
			dfs(matrix, pacific, INT_MIN, i, 0);
			dfs(matrix, atlantic, INT_MIN, i, n - 1);
		}
		for (int i = 0; i < n; ++i) {
     //行
			dfs(matrix, pacific, INT_MIN, 0, i);
			dfs(matrix, atlantic, INT_MIN, m - 1, i);
		}
		for (int i = 0; i < m; ++i)
			for (int j = 0; j < n; ++j)
				if (pacific[i][j] && atlantic[i][j])//都能流到
					ans.push_back({
      i, j });
		return ans;
	}
	void dfs(vector<vector<int>>& matrix, vector<vector<bool>>& visited, int pre, int x, int y) {
     
		if (x < 0 || x >= m || y < 0 || y >= n || visited[x][y] || matrix[x][y] < pre) return;//规定水流只能从高到低或者在同等高度上流动。
		visited[x][y] = true;
		for (int i = 0; i < 4; i++) {
     
			int nx = x + dx[i], ny = y + dy[i];
			dfs(matrix, visited, matrix[x][y], nx, ny);
		}
	}
};

方法二：广度优先搜索

思路：构造状态矩阵，第一个bit表示太平洋能否到达，第二个bit存储大西洋能否到达，11表示两个大洋都能到达

class Solution {
     
public:
    int dx[4] = {
      1,0,-1,0 }, dy[4] = {
      0,1,0,-1 };
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& matrix) {
     
        vector<vector<int> > res;
        if (matrix.empty()) return res;
        int m = matrix.size(), n = matrix[0].size();
        vector<vector<int> > status(m, vector<int>(n, 0));//m*n二维向量初始化为0
        queue<pair<int, int> > q;
        for (int i = 0; i < m; ++i) {
     
            q.push({
      i, 0 }); status[i][0] |= 1;//入队，更新状态矩阵
            q.push({
      i, n - 1 }); status[i][n - 1] |= 2;
        }
        for (int i = 0; i < n; ++i) {
     
            q.push({
      0, i }); status[0][i] |= 1;
            q.push({
      m - 1, i }); status[m - 1][i] |= 2;
        }
        while (!q.empty()) {
     //bfs
            auto p = q.front(); q.pop();
            for (int i = 0; i < 4; ++i) {
     
                int nx = p.first + dx[i], ny = p.second + dy[i];
                if (0 <= nx && nx < m && 0 <= ny && ny < n && matrix[nx][ny] >= matrix[p.first][p.second])
                    if (status[nx][ny] != status[p.first][p.second]) {
     
                        status[nx][ny] |= status[p.first][p.second];
                        q.push({
      nx, ny });
                    }
            }
        }
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (status[i][j] == 3)//11表示两个大洋都能到达
                    res.push_back({
      i, j });
        return res;
    }
};