标签：style blog http io color ar os sp for
Problem Statement (link):
‘X‘ and ‘O‘, capture all regions surrounded by ‘X‘.‘O‘s into ‘X‘s in that surrounded region.X X X X X O O X X X O X X O X X
X X X X X X X X X X X X X O X X
Analysis:
Rather than recording the 2D positions for any scanned ‘O‘, a trick is to substitute any border ‘O‘s with another character - here in the Code I use ‘Y‘. And scan the board again to change any rest ‘O‘s to ‘X‘s, and change ‘Y‘s back to ‘O‘s.
We start searching ‘O‘ from the four borders. I tried DFS first, the OJ gives Runtime error on the 250x250 large board; In the Sol 2 below, I implement BFS instead, and passed all tests.
The time complexity is O(n^2), as in the worst case, we may need to scan the entire board.
Code:
1, DFS
1 class Solution { 2 public: 3 // dfs - Runtime error on large board 250x250 4 void dfs(vector<vector<char>> &board, int r, int c) { 5 if (r<0||r>board.size()-1||c<0||c>board[0].size()-1||board[r][c]!=‘O‘) 6 return; 7 board[r][c]=‘Y‘; 8 dfs(board, r-1, c); 9 dfs(board, r+1, c); 10 dfs(board, r, c-1); 11 dfs(board, r, c+1); 12 } 13 void solve(vector<vector<char>> &board) { 14 if (board.empty() || board.size()<3 || board[0].size()<3) 15 return; 16 int r=board.size(); 17 int c=board[0].size(); 18 // dfs from boundary to inside 19 for (int i=0; i<c; i++) { 20 if (board[0][i]==‘O‘) 21 dfs(board, 0, i); // first row 22 if (board[c-1][i]==‘O‘) 23 dfs(board, c-1, i); // last row 24 } 25 for (int i=0; i<board.size(); i++) { 26 if (board[i][0]==‘O‘) 27 dfs(board, i, 0); // first col 28 if (board[i][c-1]) 29 dfs(board, i, c-1); // last col 30 } 31 // scan entire matrix and set values 32 for (int i=0; i<board.size(); i++) { 33 for (int j=0; j<board[0].size(); j++) { 34 if (board[i][j]==‘O‘) 35 board[i][j]=‘X‘; 36 else if (board[i][j]==‘Y‘) 37 board[i][j]=‘O‘; 38 } 39 } 40 } 41 };
2, BFS
1 class Solution { 2 public: 3 void solve(vector<vector<char>> &board) { 4 if (board.empty() || board.size()<3 || board[0].size()<3) 5 return; 6 int r=board.size(); 7 int c=board[0].size(); 8 // queues to store row and col indices 9 queue<int> qr; 10 queue<int> qc; 11 // start from boundary 12 for (int i=0; i<c; i++) { 13 if (board[0][i]==‘O‘) { qr.push(0); qc.push(i); } 14 if (board[r-1][i]==‘O‘) { qr.push(r-1); qc.push(i); } 15 } 16 for (int i=0; i<r; i++) { 17 if (board[i][0]==‘O‘) { qr.push(i); qc.push(0); } 18 if (board[i][c-1]==‘O‘) { qr.push(i); qc.push(c-1); } 19 } 20 // BFS 21 while (!qr.empty()) { 22 int rt=qr.front(); qr.pop(); 23 int ct=qc.front(); qc.pop(); 24 board[rt][ct]=‘Y‘; 25 if (rt-1>=0 && board[rt-1][ct]==‘O‘) { qr.push(rt-1); qc.push(ct); } //go up 26 if (rt+1<r && board[rt+1][ct]==‘O‘) { qr.push(rt+1); qc.push(ct); } // go down 27 if (ct-1>=0 && board[rt][ct-1]==‘O‘) { qr.push(rt); qc.push(ct-1); } // go left 28 if (ct+1<c && board[rt][ct+1]==‘O‘) { qr.push(rt); qc.push(ct+1); } // go right 29 } 30 31 // scan entire matrix and set values 32 for (int i=0; i<board.size(); i++) { 33 for (int j=0; j<board[0].size(); j++) { 34 if (board[i][j]==‘O‘) board[i][j]=‘X‘; 35 else if (board[i][j]==‘Y‘) board[i][j]=‘O‘; 36 } 37 } 38 } 39 };
http://justcodings.blogspot.com/2014/07/leetcode-surrounded-regions.html
leetcode-surrounded regions-ZZ
标签：style blog http io color ar os sp for
原文地址：http://www.cnblogs.com/forcheryl/p/4082004.html
