[LeetCode] Surrounded Regions

最新推荐文章于 2020-08-11 10:37:01 发布

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最新推荐文章于 2020-08-11 10:37:01 发布

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分类专栏： LeetCode

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本文介绍了一种算法，用于解决二维矩阵中被X包围的O区域的问题。该算法通过深度优先搜索确定边界上的O是否可以保持不变，并将内部孤立的O标记为需要转换成X。

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

一个二维矩阵，知道到所有连通到边缘的O,这些O是不可改变的，其余的O全部转化为X

class Solution {
public: 
    bool canChange[1001][1001];
    bool walked[1001][1001];
    void solve(vector<vector<char>> &board) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
       
        if(board.size() == 0) return ;
        vector<int> row,col;
        
        for(int i = 0; i < board.size(); i++) {
           
            for(int j = 0; j < board[i].size(); j++) {
                
                if(board[i][j] == 'X') {
                    canChange[i][j] = false;
                    walked[i][j] = true;
                }
                else {
                    canChange[i][j] = true;
                    walked[i][j] = false;
                }
                if(i == 0 || i == board.size()-1 || j == 0 || j == board[i].size()-1) {
                    if(board[i][j] == 'O') {
                        canChange[i][j] = false;
                        row.push_back(i);
                        col.push_back(j);
                    }
                }
            }
        }
        
        for(int i = 0 ; i< row.size(); i++) {
            int r = row[i];
            int c = col[i];
            turn(r,c,board);
        }        
        
        for(int i = 0; i < board.size(); i++) {
            for(int j = 0; j < board[i].size(); j++) {
                if(board[i][j] == 'O' && canChange[i][j])
                    board[i][j] = 'X';
            }
        }
    }
    
    void turn(int r, int c, vector<vector<char>> &board) {
        if(r < 0 || r >= board.size()) return ;
        if(c < 0 || c >= board[r].size()) return ;
        
        if(walked[r][c]) return;
        
        walked[r][c] = true;
        canChange[r][c] = false;
        turn(r-1,c,board);
        turn(r+1,c,board);
        turn(r,c-1,board);
        turn(r,c+1,board);
    }
    
    
};