1. 图像渲染

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
    int oldcolor; 
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        oldcolor = image[sr][sc]; // 保存原始像素值
        m = image.size();
        n = image[0].size();
        if(image[sr][sc] == color)
            return image;
        // 先将起点位置修改为color
        image[sr][sc] = color;
        dfs(image, sr, sc, color);
        return image;
    }

    void dfs(vector<vector<int>>& image, int sr, int sc, int color)
    {
        for(int k = 0; k < 4; k++)
        {
            int x = dx[k] + sr;
            int y = dy[k] + sc;
            if(x >= 0 && x < m && y >= 0 && y < n && image[x][y] == oldcolor)
            {
                image[x][y] = color;
                dfs(image, x ,y , color);
            }
        }
    }
};

 2. 岛屿数量

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    bool vis[301][301] = { false };
    int m, n;
public:
    int numIslands(vector<vector<char>>& grid) {
        int ret = 0;
        m = grid.size();
        n = grid[0].size();
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                // 如果陆地没有标记并且为1,统计结果
                if(!vis[i][j] && grid[i][j] == '1') 
                {
                    ret++;
                    vis[i][j] = true;
                    dfs(grid, i, j);  // 把这块岛屿相连的陆地全都标记  
                }
        return ret;         
    }

    void dfs(vector<vector<char>>& grid, int i, int j)
    {
        for(int k = 0; k < 4; k++)
        {
            int x = dx[k] + i;
            int y = dy[k] + j;
            if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[i][j] == '1')
            {
                vis[i][j] = true;
                dfs(grid, x ,y);
            }
        }
    }
};

3. 岛屿的最大面积

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    bool vis[51][51] = { false };
    int m, n;
    int count = 0, ret = 0;
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();
        
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                // 如果陆地没有标记并且为1,此时开始深搜
                if(!vis[i][j] && grid[i][j] == 1) 
                {
                    count = 0;
                    dfs(grid, i, j);  // 把这块岛屿相连的陆地全都标记并统计面积 
                    ret = max(ret, count);
                }
        return ret;
    }

    void dfs(vector<vector<int>>& grid, int i, int j)
    {
        count++;
        vis[i][j] = true;
        for(int k = 0; k < 4; k++)
        {
            int x = dx[k] + i;
            int y = dy[k] + j;
            if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && grid[x][y] == 1)
            {
                dfs(grid, x ,y);
            }
        }
    }
};

4. 被围绕的区域

 

正难则反，我们可以四周开始向里面遍历，但凡能深搜到的，都是不能修改的，此时我们将遍历到的标记一下，其余的全部修改成x即可。

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
public:
    void solve(vector<vector<char>>& board) {
        m = board.size();
        n = board[0].size();
        
        // 1. 把边界的O相连的联通块，全部修改成.
        // 修改两行
        for(int j = 0; j < n; j++)
        {
            if(board[0][j] == 'O') dfs(board, 0, j);
            if(board[m - 1][j] == 'O') dfs(board, m - 1, j);
        }
        // 修改两列
        for(int i = 0; i < m; i++)
        {
            if(board[i][0] == 'O') dfs(board, i, 0);
            if(board[i][n - 1] == 'O') dfs(board, i, n - 1);
        }

        // 2. 还原
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
            {
                if(board[i][j] == '.') board[i][j] = 'O';
                else if(board[i][j] == 'O') board[i][j] = 'X';
            }           
    }

     void dfs(vector<vector<char>>& board, int i, int j)
    {
        board[i][j] = '.';
        for(int k = 0; k < 4; k++)
        {
            int x = dx[k] + i;
            int y = dy[k] + j;
            if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
            {
                dfs(board, x ,y);
            }
        }
    }
};

 5. 太平洋大西洋水流问题

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
public:
    vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights) {
        m = heights.size();
        n = heights[0].size();

        vector<vector<bool>> pac(m, vector<bool>(n));
        vector<vector<bool>> atl(m, vector<bool>(n));
        
        // 先处理第一行和第一列 - 太平洋
        for(int j = 0; j < n; j++) dfs(heights, 0, j, pac);
        for(int i = 0; i < m; i++) dfs(heights, i, 0, pac);

        // 再处理最后一行和最后一列 - 大西洋
        for(int j = 0; j < n; j++) dfs(heights, m - 1, j, atl);
        for(int i = 0; i < m; i++) dfs(heights, i, n - 1, atl);

        vector<vector<int>> ret;
        for(int i = 0; i < m; i++)
            for(int j = 0; j < n; j++)
                if(atl[i][j] && pac[i][j])
                    ret.push_back({i, j});
        return ret;
    }

    void dfs(vector<vector<int>>& heights, int i, int j, vector<vector<bool>>& vis)
    {
        vis[i][j] = true;
        for(int k = 0; k < 4; k++)
        {
            int x = dx[k] + i;
            int y = dy[k] + j;
            if(x >= 0 && x < m && y >= 0 && y < n && !vis[x][y] && heights[i][j] <= heights[x][y])
            {
                dfs(heights, x, y, vis);
            }
        }
    }
};

6. 扫雷游戏

class Solution {
    int dx[8] = {0, 0, 1, -1, 1, 1, -1, -1};
    int dy[8] = {1, -1, 0, 0, 1, -1 ,1, -1};
    int m, n;
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        m = board.size();
        n = board[0].size();
        // 如果运气背,起始位置就是地雷,直接返回
        int x = click[0];
        int y = click[1];
        if(board[x][y] == 'M')
        {
            board[x][y] = 'X';
            return board;
        }
        dfs(board, x, y);
        return board;
    }

    void dfs(vector<vector<char>>& board, int i, int j)
    {
        // 先统计一下周围地雷的个数
        int count = 0;
        for(int k = 0; k < 8; k++)
        {
            int x = dx[k] + i;
            int y = dy[k] + j;
            // M 表示地雷
            if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M')
            {
                count++;
            }
        }
        // 该位置处存在地雷
        if(count)
        {
            // 存在地雷就不需要展开
            board[i][j] = count + '0';
            return;
        }
        else
        {   // 周围八个位置都要展开
            // 先把自己标记成空方块
            board[i][j] = 'B';
            for(int k = 0; k < 8; k++)
            {
                int x = dx[k] + i;
                int y = dy[k] + j;
                // E 表示没有点击的位置
                if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E')
                {
                    dfs(board, x, y);
                }
            }
        }
    }
};

7. 机器人的运动范围

class Solution {
    int dx[4] = {0, 0, -1, 1};
    int dy[4] = {-1, 1, 0, 0};
    int ret;
    bool vis[101][101] = { false };
public:
    int wardrobeFinishing(int m, int n, int cnt) {
        dfs(0, 0, m, n, cnt);
        return ret;
    }

    bool check(int i, int j, int cnt)
    {
        // 求数位之和
        int tmp = 0;
        while(i)
        {
            tmp += i % 10;
            i /= 10;
        }
        while(j)
        {
            tmp += j % 10;
            j /= 10;
        }
        return tmp <= cnt;
    }

    void dfs(int i, int j, int m, int n, int cnt)
    {   
        ret++;
        vis[i][j] = true;
        for(int k = 0; k < 4; k++)
        {
            int x = i + dx[k];
            int y = j  +dy[k];
            if(x >= 0 && x < m && y >=0 && y < n && !vis[x][y] && check(x,y,cnt))
            {

                dfs(x, y, m, n, cnt);
            }
        }
    }
};