题目原型:
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:
这个题的意思就是类似于,给你一张围棋图,如果‘O’表示白子,'X'表示黑子,问,哪些白子被围死了。自然我们想到了深度优先和广度优先搜索,那么从哪里开始搜索了,当然是从棋盘的四周了,搜索完后把被围死的白子变为黑子即可。
关于深度和广度优先搜索算法网上有很多资料,大体上可以分为两种,一种是递归另一种是非递归,在这里我们用的是非递归做法。
//广度优先搜索
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);
}
}
//深度优先搜索
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);
}
//后期处理
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;//行长
int lenOfcol = board[0].length;//列长
//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++)//此处要特别注意low的起点和终点
{
/*bfs(board, low, 0);
bfs(board, low, lenOfcol-1);*/
dfs(board, low, 0);
dfs(board, low, lenOfcol-1);
}
doit(board);
}