[leetcode] 130. Surrounded Regions 解题报告

来源：互联网发布：early stopping算法编辑：程序博客网时间：2024/06/03 16:19

题目链接：https://leetcode.com/problems/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 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

思路：如果直接从一个'O'开始搜索判断是不是会到达边界，将会非常麻烦，因为需要保存路径上的每个'O'，如果到达边界就放弃这片区域的所有值，否则将其变为'X'。有一种更为聪明的做法是从四个边界出发，用DFS算法将从边界开始的'O'的区域都变成另外一个临时的值，这样做完之后剩下的‘O’将会是被包围的，然后将其变为‘X’，再将临时的值变为'O'即可。简单粗暴！

代码如下：

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

参考：https://leetcode.com/discuss/42445/a-really-simple-and-readable-c-solution%EF%BC%8Conly-cost-12ms

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