算法：FloodFill算法

算法原理

FLoodFill算法通俗来讲，就是洪水给地势带来的变化，而实际上题目要求的就是一个连通块问题，那本质还是暴搜和DFS/BFS相结合，下面用例题来解释

图像渲染

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

    void dfs(vector<vector<int>>& image, int i, int j)
    {
        if(image[i][j] != oldcolor)
        {
            return;
        }

        image[i][j] = newcolor;
        for(int k = 0; k < 4; k++)
        {
            int x = i + dx[k], y = j + dy[k];
            if(x >= 0 && x < image.size() && y >= 0 && y < image[0].size() && image[x][y] == oldcolor)
            {
                dfs(image, x, y);
            }
        }
    }

    vector<vector<int>> floodFill(vector<vector<int>>& image, int sr, int sc, int color) 
    {
        oldcolor = image[sr][sc];
        newcolor = color;
        if(newcolor == oldcolor)
            return image;
        dfs(image, sr, sc);
        return image;
    }
};
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岛屿数量

难度不大的题，直接暴力搜索即可

class Solution 
{
public:
    // 全局变量
    bool check[301][301];
    int ret;
    int m,n;
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};

    int numIslands(vector<vector<char>>& grid) 
    {
        m = grid.size();
        n = grid[0].size();
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(grid[i][j] == '1' && check[i][j] == false)
                {
                    ret++;
                    dfs(grid, i, j);
                }
            }
        }
        return ret;
    }

    void dfs(vector<vector<char>>& grid, int i, int j)
    {
        check[i][j] = true;

        for(int k = 0; k < 4; k++)
        {
            int x = i + dx[k], y = j + dy[k];
            if(x >= 0 && x < m && y >= 0 && y < n && grid[x][y] == '1' && check[x][y] == false)
            {
                dfs(grid, x, y);
            }
        }
    }
};
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岛屿的最大面积

经验积累：

一开始把path放到了函数中，每次path+1，但测试用例过不了，经过很长时间的调试后发现了问题的所在，问题在于是在搜索后如果回溯了，以前加过的path就会减掉，所以测试用例过不了，把path放到全局变量，每次统计完后统一进行恢复就不会出现这样的情况了

class Solution 
{
public:
    bool check[51][51];
    int ret;
    int dx[4] = {0, 0, 1, -1};
    int dy[4] = {1, -1, 0, 0};
    int m, n;
    int path;

    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++)
            {
                if(grid[i][j] == 1 && check[i][j] == false)
                {
                    path++;
                    dfs(grid, i, j);
                    path = 0;
                }
            }
        }
        return ret;
    }

    void dfs(vector<vector<int>>& grid, int i, int j)
    {
        ret = max(ret, path);
        check[i][j] = true;

        for(int k = 0; k < 4; k++)
        {
            int x = i + dx[k], y = j + dy[k];
            if(x >= 0 && x < m && y >= 0 && y < n && check[x][y] == false && grid[x][y] == 1)
            {
                path++;
                dfs(grid, x, y);
            }
        }
    }
};
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被围绕的区域

经验积累：

本题采用的是正难则反的思想

class Solution 
{
public:
    // 全局变量
    int m, n;
    int dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};

    void solve(vector<vector<char>>& board) 
    {
        m = board.size();
        n = board[0].size();
        // 处理边界
        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);
        }
        for(int i = 0; i < n; i++)
        {
            if(board[0][i] == 'O')
                dfs(board, 0, i);
            if(board[m - 1][i] == 'O')
                dfs(board, m - 1, i);
        }
        for(int i = 0; i < m; i++)
        {
            for(int j = 0; j < n; j++)
            {
                if(board[i][j] == 'O')
                    board[i][j] = 'X';
                if(board[i][j] == '.')
                    board[i][j] = 'O';
            }
        }
    }

    void dfs(vector<vector<char>>& board, int i, int j)
    {
        board[i][j] = '.';

        for(int k = 0; k < 4; k++)
        {
            int x = i + dx[k], y = j + dy[k];
            if(x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O')
            {
                dfs(board, x, y);
            }
        }
    }
};
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太平洋大西洋水流问题

class Solution
{
public:
	int m, n;
	bool check1[201][201];
	bool check2[201][201];
	int dx[4] = { 1, -1, 0, 0 };
	int dy[4] = { 0, 0, 1, -1 };

	vector<vector<int>> pacificAtlantic(vector<vector<int>>& heights)
	{
		vector<vector<int>> ret;
		m = heights.size();
		n = heights[0].size();

		for (int i = 0; i < n; i++)
		{
			dfs1(heights, 0, i);
			dfs2(heights, m - 1, i);
		}

		for (int i = 0; i < m; i++)
		{
			dfs1(heights, i, 0);
			dfs2(heights, i, n - 1);
		}

		for (int i = 0; i < m; i++)
		{
			for (int j = 0; j < n; j++)
			{
				if (check1[i][j] ==true && check2[i][j] == true)
				{
					ret.push_back({ i, j });
				}
			}
		}
		return ret;
	}

	void dfs1(vector<vector<int>>& board, int i, int j)
	{
		check1[i][j] = true;
		for (int k = 0; k < 4; k++)
		{
			int x = i + dx[k], y = j + dy[k];
			if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] >= board[i][j] && check1[x][y] == false)
			{
				dfs1(board, x, y);
			}
		}
	}

	void dfs2(vector<vector<int>>& board, int i, int j)
	{
		check2[i][j] = true;
		for (int k = 0; k < 4; k++)
		{
			int x = i + dx[k], y = j + dy[k];
			if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] >= board[i][j] && check2[x][y] == false)
			{
				dfs2(board, x, y);
			}
		}
	}
};
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扫雷游戏

class Solution
{
public:
	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;
	bool check[51][51];

	vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click)
	{
		int i = click[0], j = click[1];
		m = board.size(), n = board[0].size();

		// 挖到地雷
		if (board[i][j] == 'M')
		{
			board[i][j] = 'X';
			return board;
		}

		// 挖到空方块，开始递归
		dfs(board, i, j);
		return board;
	}

	void dfs(vector<vector<char>>& board, int i, int j)
	{
		int count = 0;
		for (int k = 0; k < 8; k++)
		{
			int x = i + dx[k], y = j + dy[k];
			if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M' && check[x][y] == false)
			{
				// 统计一下四周雷的个数
				count++;
			}
		}

		if (count)
		{
			board[i][j] = count + '0';
			return;
		}
		else
		{
			board[i][j] = 'B';
			for (int k = 0; k < 8; k++)
			{
				int x = i + dx[k], y = j + dy[k];
				if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E')
				{
					// 如果周围的格子不是雷，就递归
					dfs(board, x, y);
				}
			}
		}
	}
};
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衣橱整理

以一道简单题收工~

class Solution 
{
public:
    int dx[2] = {0, 1};
    int dy[2] = {1, 0};
    int ret;
    int m, n, cnt;
    bool check[101][101];

    int func(int m)
    {
        int count = 0;
        while(m)
        {
            count += m % 10;
            m = m / 10;
        }
        return count;
    }

    int wardrobeFinishing(int _m, int _n, int _cnt) 
    {
        m = _m;
        n = _n;
        cnt = _cnt;
        dfs(0, 0);
        return ret;
    }

    void dfs(int i, int j)
    {
        check[i][j] = true;
        ret++;
        for(int k = 0; k < 2; k++)
        {
            int x = i + dx[k], y = j + dy[k];
            if(x < m && y < n && check[x][y] == false && func(x) + func(y) <= cnt)
            {
                dfs(x, y);
            }
        }
    }
};
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