Surrounded Regions leetcode java
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题目:
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
题解:
这道题是说一个O周围都是X那么这个O就得变成X。那么就可以发现四周这一圈如果有O肯定不能四周都被X包围,同时这个O也将会是潜在的内部的O的缺口,
让内部的O不能都被X覆盖。因此,思路就是先对四周的O进行特殊处理,用BFS走,把所有这个O连接的O(包括这个O)都涂成#。这样子,对于原来的棋盘来说,
没有变成#的O就是四周没有被O污染的,四周都是X,那么对其变成X。而所有#就是那些原来是O但是不是四周都被X包围的,把它们再还原成O。代码如下:
public void solve(char[][] board) { if(board==null || board.length<=1 || board[0].length<=1) return; //第一行和最后一行进行fill for(int i=0;i<board[0].length;i++){ fill(board,0,i); fill(board,board.length-1,i); } //第一列和最后一列进行fill for(int i=0;i<board.length;i++){ fill(board,i,0); fill(board,i,board[0].length-1); } //对于当前格子中,所有O变成X(说明符合要求,没有被变成#),所有#变成O for(int i=0;i<board.length;i++){ for(int j=0;j<board[0].length;j++){ if(board[i][j]=='O') board[i][j]='X'; else if(board[i][j]=='#') board[i][j]='O'; } }}private void fill(char[][] board, int i, int j){ if(board[i][j]!='O') return; board[i][j] = '#'; //利用BFS LinkedList<Integer> queue = new LinkedList<Integer>(); //先将矩阵的横纵坐标编码 int code = i*board[0].length+j; queue.add(code); while(!queue.isEmpty()){ code = queue.poll(); int row = code/board[0].length;//从code中还原横坐标 int col = code%board[0].length;//从code中还原纵坐标 if(row>=1 && board[row-1][col]=='O'){//当前元素上边是否为0 queue.add((row-1)*board[0].length + col); board[row-1][col]='#'; } if(row<=board.length-2 && board[row+1][col]=='O'){//当前元素下面是否为0 queue.add((row+1)*board[0].length + col); board[row+1][col]='#'; } if(col>=1 && board[row][col-1]=='O'){//当前元素左边是否为0 queue.add(row*board[0].length + col-1); board[row][col-1]='#'; } if(col<=board[0].length-2 && board[row][col+1]=='O'){//当前元素右边是否为0 queue.add(row*board[0].length + col+1); board[row][col+1]='#'; } }}Reference:http://www.cnblogs.com/springfor/p/3869853.html
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