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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
思路：
反向思考，先找到没有被包围的区域，标记为‘+’,再把标为‘+’的区域标为‘O‘,标为’O‘的区域改为‘X‘。没有被包围的区域一定是与最外圈的’O‘相连的区域，所以要先遍历区域的上下左右边界，找到‘O‘的地方。接下来的问题是如何把所有与最外层‘O‘相连的区域标记上。有两种思路：BFS和DFS。
所谓BFS是指把当前标记位置的上下左右都标记一遍，然后再标记相邻点的上下左右位置。不需递归，用队列。
DFS是指把当前标记位置的向一个方向标记，比如一直向左，直到没有可标记的，再换一个方向。需要递归。
代码里面BFS可以通过，DFS栈溢出了。
void solve(vector<vector<char>> &board) { if(board.size() == 0) return; int rowNum = board.size(); int colNum = board[0].size(); //遍历最外面一圈，找‘O‘ //最上 for(int j = 0; j < colNum; j++) { if(board[0][j] == ‘O‘) BFS(board, 0, j); } //最下 for(int j = 0; j < colNum; j++) { if(board[rowNum - 1][j] == ‘O‘) BFS(board, rowNum - 1, j); } //最左 for(int i = 0; i < rowNum; i++) { if(board[i][0] == ‘O‘) BFS(board, i, 0); } //最右 for(int i = 0; i < rowNum; i++) { if(board[i][colNum - 1] == ‘O‘) BFS(board, i, colNum - 1); } for(int i = 0; i < rowNum; i++) { for(int j = 0; j < colNum; j++) { if(board[i][j] == ‘O‘) board[i][j] = ‘X‘; if(board[i][j] == ‘+‘) board[i][j] = ‘O‘; } } } void DFS(vector<vector<char>> &board, int r, int c) { if(r >= 0 && c >= 0 && r < board.size() && c < board[0].size() && board[r][c] == ‘O‘) { board[r][c] = ‘+‘; DFS(board, r - 1, c); DFS(board, r + 1, c); DFS(board, r, c - 1); DFS(board, r, c + 1); } } void BFS(vector<vector<char>> &board, int r, int c) { queue<pair<int, int>> q; q.push(make_pair(r, c)); while(!q.empty()) { int i = q.front().first; int j = q.front().second; q.pop(); if(i >= 0 && j >= 0 && i < board.size() && j < board[0].size() && board[i][j] == ‘O‘) { board[i][j] = ‘+‘; q.push(make_pair(i - 1, j)); q.push(make_pair(i + 1, j)); q.push(make_pair(i, j - 1)); q.push(make_pair(i, j + 1)); } } }
上面的代码已经是优化过的了，我自己写的时候只写出了DFS的，而且自己也没有意识到是DFS。代码也很繁琐。注意通过把判断条件放在一起来简化代码。
我原本很挫的代码：栈溢出。
class Solution { public: void solve(vector<vector<char>> &board) { if(board.size() == 0) return; int rowNum = board.size(); int colNum = board[0].size(); //遍历最外面一圈，找‘O‘ //最上 for(int j = 0; j < colNum; j++) { if(board[0][j] == ‘O‘) { board[0][j] = ‘+‘; mySolve(board, 0, j); } } //最下 for(int j = 0; j < colNum; j++) { if(board[rowNum - 1][j] == ‘O‘) { board[rowNum - 1][j] = ‘+‘; mySolve(board, rowNum - 1, j); } } //最左 for(int i = 0; i < rowNum; i++) { if(board[i][0] == ‘O‘) { board[i][0] = ‘+‘; mySolve(board, i, 0); } } //最右 for(int i = 0; i < rowNum; i++) { if(board[i][colNum - 1] == ‘O‘) { board[i][colNum - 1] = ‘+‘; mySolve(board, i, colNum - 1); } } for(int i = 0; i < rowNum; i++) { for(int j = 0; j < colNum; j++) { if(board[i][j] == ‘O‘) board[i][j] = ‘X‘; if(board[i][j] == ‘+‘) board[i][j] = ‘O‘; } } } void mySolve(vector<vector<char>> &board, int r, int c) { if(r - 1 >= 0 && board[r - 1][c] == ‘O‘) //上 { board[r - 1][c] == ‘+‘; mySolve(board, r - 1, c); } if(r + 1 < board.size() && board[r + 1][c] == ‘O‘) //下 { board[r + 1][c] == ‘+‘; mySolve(board, r + 1, c); } if(c - 1 >= 0 && board[r][c - 1] == ‘O‘) //左 { board[r][c - 1] == ‘+‘; mySolve(board, r, c - 1); } if(c + 1 < board[0].size() && board[r][c + 1] == ‘O‘) //下 { board[r][c + 1] == ‘+‘; mySolve(board, r, c + 1); } } };
【leetcode】Surrounded Regions（middle）☆
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原文地址：http://www.cnblogs.com/dplearning/p/4474768.html
