【LeetCode】Surrounded Regions

最新推荐文章于 2020-08-11 10:37:01 发布

Snowwolf_Yang

最新推荐文章于 2020-08-11 10:37:01 发布

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本文链接：https://blog.youkuaiyun.com/snowwolf_yang/article/details/26471125

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参考链接

http://blog.youkuaiyun.com/doc_sgl/article/details/11836967
http://blog.youkuaiyun.com/pickless/article/details/12074363

题目描述

Surrounded Regions

Given a 2D board containing 'X' and 'O', capture all regions surrounded by 'X'.

A region is captured by flipping all 'O's into 'X's in that surrounded region.

For example,

X X X X
X O O X
X X O X
X O X X

After running your function, the board should be:

X X X X
X X X X
X X X X
X O X X

题目分析

从四个边界的'O'出发，它们所能到达的'O'就是没有被包围的'O'。

所以，该题可以用BFS遍历或者DFS遍历。

代码示例

<pre name="code" class="cpp">//深度优先遍历，超时 

class Solution {
public:
    void solve(vector<vector<char> > &board) {
    	int m = board.size();
		if(m < 1)	return;
		int n = board[0].size();
		//把与四边上的O相连的所有O换为# 
		for(int i = 0;i<m;i++)//左 
			solveCore(board,i,0);
		for(int i = 0;i<m;i++)//右 
			solveCore(board,i,n-1);
			
		for(int j = 1;j<n-1;j++)//上 
			solveCore(board,0,j);
		for(int j = 1;j<n-1;j++)//下 
			solveCore(board,m-1,j);
		
		//把所有O换为X 
		for(int i = 0;i<m;i++)
		{
			for(int j = 0;j<n;j++)
			{
				if(board[i][j] == 'O')
					board[i][j] = 'X';
			}
		}
		//把所有#换为O 
		for(int i = 0;i<m;i++)
		{
			for(int j = 0;j<n;j++)
			{
				if(board[i][j] == '#')
					board[i][j] = 'O';
			}
		}
		    
    }
    void solveCore(vector<vector<char> > &board,int i, int j)
    {
    	if(i < 0 || j < 0 || i >= board.size() || j >= board[0].size() || board[i][j] != 'O')
			return;
	
		//printf("%d-%d:\n",i,j);
		//printvecvec(board,"core");	
		//system("pause");
		board[i][j] = '#';
		
		solveCore(board,i,j-1);
		solveCore(board,i+1,j);
		solveCore(board,i,j+1);		
		solveCore(board,i-1,j);
	}
};

//广度优先遍历，通过 

class Solution {
public:
    queue<int> q;
    int m, n;
    
    void add(int x, int y, vector<vector<char>> &board) {
        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O') {
            board[x][y] = 'Z';
            q.push(x * n + y);
        }    
    }
    
    void traversal(int x, int y, vector<vector<char>> &board) {
        add(x, y, board);
        while (!q.empty()) {
            int p = q.front();
            q.pop();            
            int px = p / n, py = p % n;
            add(px - 1, py, board);
            add(px + 1, py, board);
            add(px, py - 1, board);
            add(px, py + 1, board);
        }
    }
    
    void solve(vector<vector<char>> &board) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (board.empty() || board.size() == 0 || board[0].size() == 0) {
            return;
        }
        m = board.size(), n = board[0].size();
        for (int i = 0; i < n; i++) {
            traversal(0, i, board);
            traversal(m - 1, i, board);
        }
        for (int i = 0; i < m; i++) {
            traversal(i, 0, board);
            traversal(i, n - 1, board);
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                board[i][j] = board[i][j] == 'Z' ? 'O' : 'X';
            }
        }
    }
};