130. Surrounded Regions

问题描述 Given a 2D board containing ‘X’ and ‘O’ (the letter 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

解决思路 对边缘的元素进行搜索，将边缘的o元素，以及与边缘o元素相邻的o元素变成‘w‘存储。然后将剩下的o元素变成x，将w重新变回o即可。值得注意的是，这里需要考虑空间问题，要进行枝，搜索过的就不要再搜索了。。。

代码

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