130. Surrounded Regions
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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 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
题意:
将所有被X包住的O都变成X,但是和边缘相邻的O不算被包围
思路:
DFS:只从上下左右四个边缘开始递归,将遇到的‘O’全部修改为一个特殊的字符例如‘$’,最后遍历一遍数组,将‘$’改为‘O’,非‘$’的位置改为‘X’。
注:
- 只有元素为‘O’时才开始递归
- 对于递归终止的条件,最好在进入递归前用if限制,而不是在进入后用return终止。
下面两种实现方式相差80ms左右,二他们唯一的区别只是定义递归是用的dir的位置不一样,92ms的在dfs函数中定义。12ms的定义为全局变量。因为如果定义在dfs函数中,每次递归调用dfs都要初始化一次dir。这种在递归函数中使用的变量,尽量定义为全局的。不要每次调用时都要重新声明初始化。
解法1:92ms
class Solution {public: void solve(vector<vector<char>>& board) { if (board.empty()) return; int m = board.size(), n = board[0].size(); for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (i == 0 || i == m - 1 || j == 0 || j == n - 1) { if (board[i][j] == 'O') dfs(board, i , j); } } } for (int i = 0; i < m; ++i) { for (int j = 0; j < n; ++j) { if (board[i][j] == 'O') board[i][j] = 'X'; if (board[i][j] == '$') board[i][j] = 'O'; } } } void dfs(vector<vector<char>> &board, int x, int y) { int m = board.size(), n = board[0].size(); vector<vector<int>> dir{{0,-1},{-1,0},{0,1},{1,0}}; //导致耗时的原因 board[x][y] = '$'; for (int i = 0; i < dir.size(); ++i) { int dx = x + dir[i][0], dy = y + dir[i][1]; if (dx >= 0 && dx < m && dy > 0 && dy < n && board[dx][dy] == 'O') { dfs(board, dx, dy); } } }};
解法2 :12ms
int dx[] = {1,-1,0,0};int dy[] = {0,0,1,-1};class Solution {public: void solve(vector<vector<char>>& board) { if(!board.size()) return; int n = board.size(); int m = board[0].size(); for(int i = 0;i<n;++i){ for(int j = 0;j < m;++j){ if (i == 0 || i == n - 1 || j == 0 || j == m - 1) { if(board[i][j] == 'O') dfs(i,j,board); } } } for(int i = 0;i<n;++i){ for(int j = 0;j < m;++j){ if(board[i][j]=='O') board[i][j]='X'; if(board[i][j]=='$') board[i][j]='O'; } } return; } void dfs(int i,int j,vector<vector<char> >& board){ int n = board.size(), m = board[0].size(); board[i][j]='$'; for(int k = 0;k < 4;++k){ int nx = dx[k] + i; int ny = dy[k] + j; if(nx >= 0 && nx < n &&ny > 0&& ny < m && board[nx][ny] == 'O') //注意必须ny>0,ny=0会报错,这是这道题的bug dfs(nx,ny,board); } return; }};
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