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

解题思路：

　　参照网上的解题思路：先从四周的‘O'进行搜索，将能到达的'O'换成第三种符号，如（'#')，再遍历所有符号把O换成X，第三种符号换回O。

代码：

class Solution {
public:  int m, n;void traversal(int x, int y, vector<vector<char>> &board) {if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'O') {board[x][y] = 'Z';traversal(x - 1, y, board);traversal(x + 1, y, board);traversal(x, y - 1, board);traversal(x, y + 1, board);}    }void solve(vector<vector<char>> &board) {// Start typing your C/C++ solution below// DO NOT write int main() functionif (board.empty() || board.size() == 0 || board[0].size() == 0) {return;}m = board.size(), n = board[0].size();for (int i = 0; i < n; i++) {traversal(0, i, board);traversal(m - 1, i, board);}for (int i = 0; i < m; i++) {traversal(i, 0, board);traversal(i, n - 1, board);}for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {board[i][j] = board[i][j] == 'Z' ? 'O' : 'X';}}}
};

转载于:https://www.cnblogs.com/dongguangqing/p/3727285.html

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LeetCode OJ - Surrounded Regions

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