題目原型:
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
基本思路:
這個題的意思就是類似於,給你一張圍棋圖,如果‘O’表示白子,'X'表示黑子,問,哪些白子被圍死了。自然我們想到了深度優先和廣度優先搜索,那麼從哪裏開始搜索了,當然是從棋盤的四周了,搜索完後把被圍死的白子變爲黑子即可。
關於深度和廣度優先搜索算法網上有很多資料,大體上可以分爲兩種,一種是遞歸另一種是非遞歸,在這裏我們用的是非遞歸做法。
//廣度優先搜索
public void bfs(char[][] board, int low , int col)
{
int lenOfcol = board[0].length;
Queue<Integer> queue = new LinkedList<Integer>();
visit(board, low, col, queue);
while(!queue.isEmpty())
{
int current = queue.poll();
int i = current/lenOfcol;
int j = current%lenOfcol;
visit(board, i-1, j, queue);
visit(board, i, j-1, queue);
visit(board, i+1, j, queue);
visit(board, i, j+1, queue);
}
}
//深度優先搜索
public void dfs(char[][] board, int low , int col)
{
Stack<Integer> stack = new Stack<Integer>();
int lenOflow = board.length;
int lenOfcol = board[0].length;
if(!stack.contains(low*lenOfcol+col))
{
visit(board, low, col, stack);
}
while(!stack.isEmpty())
{
int current = stack.peek();
int i = current/lenOfcol;
int j = current%lenOfcol;
if(i-1>=0&&i-1<lenOflow&&board[i-1][j]=='O')
{
visit(board, i-1, j, stack);
continue;
}
if(j-1>=0&&j-1<lenOfcol&&board[i][j-1]=='O')
{
visit(board, i, j-1, stack);
continue;
}
if(i+1>=0&&i+1<lenOflow&&board[i+1][j]=='O')
{
visit(board, i+1, j, stack);
continue;
}
if(j+1>=0&&j+1<lenOfcol&&board[i][j+1]=='O')
{
visit(board, i, j+1, stack);
continue;
}
else
stack.pop();
}
}
public void visit(char[][] board, int low , int col,Stack<Integer> stack)
{
int lenOflow = board.length;
int lenOfcol = board[0].length;
if(low<0||low>=lenOflow||col<0||col>=lenOfcol||board[low][col]!='O')
return;
board[low][col] = '+';
stack.push(low*lenOfcol+col);
}
public void visit(char[][] board, int low , int col,Queue<Integer> queue)
{
int lenOflow = board.length;
int lenOfcol = board[0].length;
if(low<0||low>=lenOflow||col<0||col>=lenOfcol||board[low][col]!='O')
return;
board[low][col] = '+';
queue.offer(low*lenOfcol+col);
}
//後期處理
public void doit(char[][] board)
{
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';
if(board[i][j]=='+')
board[i][j]='O';
}
}
}
public void solve(char[][] board)
{
if(board==null||board.length==0||board[0].length==0)
return;
int lenOflow = board.length;//行長
int lenOfcol = board[0].length;//列長
//第一行,第一列,第lenOflow行,第lenOfcol列
for(int col = 0;col<lenOfcol;col++)
{
/*bfs(board, 0, col);
bfs(board, lenOflow-1, col);*/
dfs(board, 0, col);
dfs(board, lenOflow-1, col);
}
for(int low = 1;low<lenOflow-1;low++)//此處要特別注意low的起點和終點
{
/*bfs(board, low, 0);
bfs(board, low, lenOfcol-1);*/
dfs(board, low, 0);
dfs(board, low, lenOfcol-1);
}
doit(board);
}