Leetcode149: Surrounded Regions

来源：互联网发布：农村淘宝土豪村编辑：程序博客网时间：2024/06/18 05:35

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 XX O O XX X O XX O X X

After running your function, the board should be:

X X X XX X X XX X X XX O X X

class Solution {public:    void solve(vector<vector<char>>& board) {        int rows = board.size();        if(rows==0)            return;        int cols = board[0].size();        for(int i=0;i<rows;i++){            if(board[i][0]=='O'){                dfs(board,i,0);            }            if(board[i][cols-1]=='O')                dfs(board,i,cols-1);        }        for(int j=0;j<cols;j++){            if(board[0][j]=='O')                dfs(board,0,j);            if(board[rows-1][j]=='O')                dfs(board,rows-1,j);        }        for(int i=0;i<rows;i++){            for(int j=0;j<cols;j++){                if(board[i][j]=='Y')                    board[i][j]='O';                else                    board[i][j]='X';            }        }    }    void dfs(vector<vector<char>>& board, int x, int y){        int rows = board.size();        int cols = board[0].size();        board[x][y]='Y';       // cout<<x<<" "<<y<<endl;        int dir[4][2] = {0,1,1,0,0,-1,-1,0};        for(int i=0;i<4;i++){            int tx = x+dir[i][0];            int ty = y+dir[i][1];            while(tx>=0&&tx<rows&&ty>=0&&ty<cols&&board[tx][ty]=='O'){                board[tx][ty]='Y';                //cout<<tx<<" "<<ty<<endl;                for(int k=1;k<=3;k+=2){                    int px = tx+dir[(i+k)%4][0];                    int py = ty+dir[(i+k)%4][1];                    if(px>=0&&px<rows&&py>=0&&py<cols&&board[px][py]=='O')dfs(board,px,py);                }                tx = tx+dir[i][0];                ty = ty+dir[i][1];            }        }    }};

上面是看到的别人用的方法，自己用之前的dfs方法总是出现runtime error错误，心塞。

class Solution {public:    void search(int x, int y, vector<vector<char>>& board)    {        if(x<0 || x>=board.size() || y<0 || y>=board[0].size())   return;        if(board[x][y] == 'O')        {            board[x][y] = '1';            search(x, y-1, board);            search(x, y+1, board);            search(x-1, y, board);            search(x+1, y, board);        }    }    void solve(vector<vector<char>>& board) {        int m = board.size();        if(m<=2)    return;        int n = board[0].size();        if(n<=2)    return;                for(int i = 0; i < n; i++)        {            if(board[0][i] == 'O')                search(0, i, board);        }        for(int i = 0; i < n; i++)        {            if(board[m-1][i] == 'O')                search(m-1, i, board);        }        for(int i = 0; i < m-1; i++)        {            if(board[i][0] == 'O')                search(i, 0, board);        }        for(int i = 0; i < m-1; i++)        {            if(board[i][n-1] == 'O')                search(i, n-1, board);        }                for(int i = 0; i < m; i++)        {            for(int j = 0; j < n; j++)            {                if(board[i][j] == '1')                    board[i][j] = 'O';                else if(board[i][j] == 'O')                    board[i][j] = 'X';            }        }            }};

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