[leetcode]Surrounded Regions

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Surrounded Regions

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 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

演算法思路：由外向內掃描，因為邊界的O肯定是escape的，這樣由邊界往裡面縱深尋找出沒有被capture的O

思路1：dfs，第一次居然過了，然後就再也過不去了，試想一個矩陣是100* 100並且全是O，如果dfs的話，就會有10000層，棧空間都溢出了....

思路2：BFS，大資料無壓力過

代碼如下：

 1 public class Solution { 2 public void solve(char[][] board) { 3         if(board == null || board.length == 0 ) return; 4         int height = board.length; 5         int width = board[0].length; 6         int code = Math.max(height, width); 7         for(int i = 0; i < height; i++){ 8             for(int j = 0; j < width; j++){ 9                 if(board[i][j] == ‘O‘ && (i == 0 || i == height - 1 || j == 0 || j == width - 1)){10                     board[i][j] = ‘@‘;11                     bfs(board, height, width, i, j, code);12                 }13             }14         }15         for(int i = 0; i < height; i++){16             for(int j = 0; j < width; j++){17                 if(board[i][j] == ‘O‘){18                     board[i][j] = ‘X‘;19                 }else if(board[i][j] == ‘@‘){20                     board[i][j] = ‘O‘;21                 }22             }23         }24     }25     int[][] dir = {{-1,0},{1,0},{0,1},{0,-1}};26     private void bfs(char[][] board,int height,int width,int i ,int j,int code){27         Queue<Integer> q = new LinkedList<Integer>();28         q.offer(i * code + j);//將二維下標壓縮成一維，方便儲存29         while(!q.isEmpty()){30             int tem = q.poll();31             int row = tem / code;32             int col = tem % code;33             for(int k = 0;k < dir.length; k++){34                 if(row + dir[k][0] < height && row + dir[k][0] >= 0 && col + dir[k][1] < width && col + dir[k][1] >= 0){35                     if(board[row + dir[k][0]][col + dir[k][1]] == ‘O‘){36                         board[row + dir[k][0]][col + dir[k][1]] = ‘@‘;37                         q.offer((row + dir[k][0]) * code + col + dir[k][1]);38                     }39                 }40             }41         }42     }43 }

座標壓縮法很有意思。