LeetCode 130 Surrounded Regions 連通塊DFS
阿新 • • 發佈：2022-05-16
Given an m x n matrix board containing 'X' and 'O', capture all regions that are 4-directionally surrounded by 'X'.
A region is captured by flipping all 'O's into 'X's in that surrounded region.
Solution
如果由 'O' 組成的一個連通塊沒有與邊界相連， 那麼最後就會被翻轉成 'X'。因此我們可以從邊界出發進行 \(DFS\)：此時得到的連通塊都是不會被翻轉的，我們用 '#' 標記即可。最後統計時，沒有被 '#'
'O' 都是被翻轉的。
點選檢視程式碼
class Solution { private: int dx[4][2]={ 1,0, 0,1, -1,0, 0,-1 }; bool check(int i,int j,int n,int m){ if(i<0 || j<0 || i>=n || j>=m) return false; return true; } void dfs(vector<vector<char>>& board, int x, int y, int n, int m){ for(int i=0;i<4;i++){ int nx = x+dx[i][0], ny = y+dx[i][1]; if(check(nx,ny,n,m)){ if(board[nx][ny]=='O'){ board[nx][ny]='#';dfs(board,nx,ny,n,m); } } } } public: void solve(vector<vector<char>>& board) { int n = board.size(), m = board[0].size(); if(n==0)return; for(int i=0;i<n;i++){ if(board[i][0]=='O'){board[i][0]='#';dfs(board,i,0,n,m);} if(board[i][m-1]=='O'){board[i][m-1]='#';dfs(board,i,m-1,n,m);} } for(int j=0;j<m;j++){ if(board[0][j]=='O'){board[0][j]='#';dfs(board,0,j,n,m);} if(board[n-1][j]=='O'){board[n-1][j]='#';dfs(board,n-1,j,n,m);} } for(int i=0;i<n;i++){ for(int j=0;j<m;j++){ if(board[i][j]=='#')board[i][j]='O'; else if(board[i][j]=='O')board[i][j]='X'; } } } };