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 X X O O X X X O X X O X X
After running your function, the board should be:
X X X X X X X X X X X X X O X X
还是老毛病啊,整体思路一下就有了,但是写出来的代码总不能一下就过。
其实前一阵面微软的时候二面考的算法就跟这个有相似的地方,总的来说都可以归结为一个连通性的问题,用BFS;
1.很明显在最外面一圈的O挨着的O肯定最后不是被包围的,与这个O也相连的也不能被包围,即这些O在最后的矩阵里面还是O;
2.我们从最外面一圈的O出发,用BFS,利用队列帮助进行扩展,将所有它能联通到的O都标记,
3.随后我们再扫描一遍矩阵,被标记的改回O,没有被标记的O都是可以被包围的,所以要改成X;
代码:
上次面微软现场写代码,有一段跟下面 //extending 那里一样,要从四个方向扩展,然后面试官就很不满意,说你这个那不得写四次,我说要扩展四个方向,所以是要写四次,结果他非常不满,说这样的代码非常不整洁,有方法可以改的,然后我问他怎么写他又不说,各位有知道的吗,烦请告知,或者有那本书推荐的吗?
class Solution { public: void process(int i,int j,vector<vector<char> >& board) { int m=board.size(); int n=board[0].size(); typedef pair<int,int> point; queue<point> Q; Q.push(point(i,j)); board[i][j]='E'; while(!Q.empty()) { point tmp=Q.front(); Q.pop(); int x=tmp.first,y=tmp.second; //extending if (x!=0&&board[x-1][y]=='O') { Q.push(point(x-1,y)); board[x-1][y]='E'; //extended; } if (x!=m-1&&board[x+1][y]=='O') { Q.push(point(x+1,y)); board[x+1][y]='E'; //extended; } if (y!=0&&board[x][y-1]=='O') { Q.push(point(x,y-1)); board[x][y-1]='E'; //extended; } if (y!=n-1&&board[x][y+1]=='O') { Q.push(point(x,y+1)); board[x][y+1]='E'; //extended; } } } void solve(vector<vector<char> >& board) { int m=board.size(); if (m==0) return; int n=board[0].size(); int i,j; for(i=0;i<m;i++) { if (board[i][0]=='O') process(i,0,board); if (board[i][n-1]=='O') process(i,n-1,board); } for(j=0;j<n;j++) { if (board[0][j]=='O') process(0,j,board); if (board[m-1][j]=='O') process(m-1,j,board); } for(i=0;i<m;i++) for(j=0;j<n;j++) { if(board[i][j]=='O') board[i][j]='X'; else if (board[i][j]=='E') board[i][j]='O'; } } };