[LeetCode]130.Surrounded Regions

题目

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

分析

广搜。
从四个边向里面搜索，只要是遇到的’O’，都是可以从外边界走通的，都会用’.’标记
广搜一遍之后，一个一个元素遍历， 如果是’.’说明这个’O’是可以从外界走通的，保留
如果是’O’,说明这是闭塞的，从外界走不通的,替换为’X’。

代码

    /**------------------------------------
    *   日期：2015-02-06
    *   作者：SJF0115
    *   题目: 130.Surrounded Regions
    *   网址：https://oj.leetcode.com/problems/surrounded-regions/
    *   结果：AC
    *   来源：LeetCode
    *   博客：
    ---------------------------------------**/
    #include <iostream>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    using namespace std;

    class Solution {
    public:
        void solve(vector<vector<char> > &board) {
            int row = board.size();
            if(row == 0){
                return;
            }//if
            int col = board[0].size();
            // 都够不成围绕
            if(row <= 2 || col <= 2){
                return;
            }//if
            // 行
            for(int i = 0;i < col;++i){
                // 第一行
                BFS(board,row,col,0,i);
                // 最后一行
                BFS(board,row,col,row-1,i);
            }//for
            // 列
            for(int j = 0;j < row;++j){
                // 第一列
                BFS(board,row,col,j,0);
                // 最后一列
                BFS(board,row,col,j,col-1);
            }//for
            for(int i = 0;i < row;++i){
                for(int j = 0;j < col;j++){
                    // 不可以从外界走通的o
                    if(board[i][j] == 'O'){
                        board[i][j] = 'X';
                    }//if
                    // 可以从外界走通的o
                    else if(board[i][j] == '.'){
                        board[i][j] = 'O';
                    }
                }//for
            }//for
        }
    private:
        // row 行数 col 列数 x ,y 当前board位置
        void BFS(vector<vector<char>> &board,int row,int col,int x,int y){
            queue<pair<int,int> > q;
            Pass(board,row,col,x,y,q);
            while(!q.empty()){
                pair<int,int> point = q.front();
                q.pop();
                x = point.first;
                y = point.second;
                // left
                Pass(board,row,col,x,y+1,q);
                // right
                Pass(board,row,col,x,y-1,q);
                // up
                Pass(board,row,col,x-1,y,q);
                // down
                Pass(board,row,col,x+1,y,q);
            }//while
        }
        // 四边判断是否走通
        void Pass(vector<vector<char>> &board,int row,int col,int x,int y,queue<pair<int,int> > &q){
            // 边界条件以及遇到o才能走通
            if(x < 0 || x >= row || y < 0 || y >= col || board[x][y] != 'O'){
                return;
            }//if
            // 标记可从外界走通的o
            board[x][y] = '.';
            // 入队列
            q.push(make_pair(x,y));
        }
    };

    int main(){
        Solution s;
        /*vector<vector<char> > board = {
            {'X','X','X','X'},
            {'X','O','O','X'},
            {'X','X','O','X'},
            {'X','O','X','X'}
        };*/
        vector<vector<char> > board = {
            {'X','X','X'},
            {'X','O','X'},
            {'X','X','X'}
        };
        s.solve(board);
        // 输出
        for(int i = 0;i < board.size();i++){
            for(int j = 0;j < board[i].size();j++){
                cout<<board[i][j]<<" ";
            }//for
            cout<<endl;
        }//for
        return 0;
    }

运行时间

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