BFS解决FloodFill算法相关leetcode算法题

1.图像渲染

图像渲染

class Solution {
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
public:
    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) {
        int m = image.size(), n = image[0].size();
        int prev = image[sr][sc];//保存一下先前的像素值
        if(prev == color) return image;//处理一下边界情况
        queue<pair<int,int>> q;
        q.push({sr,sc});
        while(q.size())
        {
            auto& [a,b] = q.front();
            q.pop();
            image[a][b] = color;
            for(int i=0;i<4;i++)
            {
                int x = a+dx[i],y = b+dy[i];
                if(x>=0&&x<m&&y>=0&&y<n&&image[x][y]==prev)
                {
                    q.push({x,y});
                }
            }
        }
        return image;
    }
};
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2.岛屿数量

岛屿数量

class Solution {
    int dx[4] = {0,0,1,-1};
    int dy[4] = {1,-1,0,0};
    bool vis[301][301];//标志数组，用于标记某一个点是否已经访问过
    int m,n;
public:
    int numIslands(vector<vector<char>>& grid) {
       m = grid.size(), n = grid[0].size();
        int ret = 0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]=='1' && !vis[i][j])
                {
                    ret++;
                    bfs(grid,i,j);//将这一块区域都标记一下
                }
            }
        }
        return ret;
    }
    void bfs(vector<vector<char>>& grid,int i,int j)
    {
        queue<pair<int,int>> q;
        q.push({i,j});
        vis[i][j] = true;
        while(q.size())
        {
            auto [a,b] = q.front();//注意这里是拷贝不能是引用，因为引用的话，递归调用会修改a,b的值
            q.pop();
            for(int k=0;k<4;k++)
            {
                int x = a+dx[k], y = b+dy[k];
                if(x>=0 && x<m && y>=0 && y<n && grid[x][y]=='1' && !vis[x][y])
                {
                    q.push({x,y});
                    vis[x][y] = true;
                }
            }
        }
    }
};
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3.岛屿的最大面积

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];
    int m,n;
public:
    int maxAreaOfIsland(vector<vector<int>>& grid) {
        m = grid.size(),n = grid[0].size();
        int ret = 0;
        for(int i=0;i<m;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]== 1 &&!vis[i][j])
                {
                    ret = max(ret,bfs(grid,i,j));
                }
            }
        }
        return ret;
    }
    int bfs(vector<vector<int>>& grid,int i,int j)
    {
        int count = 0;
        queue<pair<int,int>> q;
        q.push({i,j});
        vis[i][j] = true;
        count++;
        while(q.size())
        {
            auto [a,b] = q.front();
            q.pop();
            for(int k=0;k<4;k++)
            {
                int x = a+dx[k],y=b+dy[k];
                if(x>=0 && x<m && y>=0 && y<n && grid[x][y]== 1 && !vis[x][y])
                {
                    count++;
                    q.push({x,y});
                    vis[x][y] = true;
                }
            }
        }
        return count;
    }
};
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4.被围绕的区域

被围绕的区域

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();
        //先处理边界上的'0'连通块
        for(int i=0;i<m;i++)
        {
            if(board[i][0] == 'O') bfs(board,i,0);
            if(board[i][n-1]=='O') bfs(board,i,n-1);
        }
        for(int j=0;j<n;j++)
        {
            if(board[0][j] == 'O') bfs(board,0,j);
            if(board[m-1][j] == 'O') bfs(board,m-1,j);
        }
        //还原
        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]=='.') board[i][j]='O';
            }
        }
    }
    void bfs(vector<vector<char>>& board,int i,int j)
    {
        queue<pair<int,int>> q;
        q.push({i,j});
        board[i][j] = '.';
        while(q.size())
        {
            auto [a,b] = q.front();
            q.pop();
            for(int k=0;k<4;k++)
            {
                int x = a+dx[k],y = b+dy[k];
                if(x>=0 && x<m && y>=0 && y<n && board[x][y]=='O')
                {
                    board[x][y] = '.';
                    q.push({x,y});
                }
            }
        }
    }
};
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