问题描述:
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
解法:题目要求将被'X'包围的字符'O'变成'X',因此这题的解法就是搜索边界上的字符'O',并找出所有与边界字符'O'连通的字符'O',将这些字符做个标记,最后再次遍历此2D board,将所有字符'O'变成'X',并将刚才做标记的字符变换为原来的'O'.所以关键是找出所有与边界上'O'字符连通的'O'字符,这是图的遍历问题,可以用DFS或BFS方法,注意用DFS遍历的时候最好要用堆栈而不是递归,我刚开始用递归时,当矩阵大于250*250时,出现了stackoverflow问题。接下来上代码:
1.递归版DFS(RunTimeError)
class Solution {
public:
void solve(vector<vector<char> > &board)
{
int m = board.size();
if (m < 3)
return;
int n = board[0].size();
if (n < 3)
return;
for (int i = 1; i < m - 1; i++)
{
dfs(board, i, 0);
dfs(board, i, n - 1);
}
for (int j = 0; j < n; j++)
{
dfs(board, 0, j);
dfs(board, m - 1, j);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 'O') board[i][j] = 'X';
else if (board[i][j] == 'A') board[i][j] = 'O';
}
}
}
private:
void dfs(vector<vector<char> > &board, int x, int y)
{
if (x<0 || x >= board.size()|| y < 0 || y >= board[0].size())
return;
if (board[x][y] == 'O')
{
board[x][y] = 'A';
dfs(board, x - 1, y);
dfs(board, x + 1, y);
dfs(board, x, y - 1);
dfs(board, x, y + 1);
}
}
};
2.用堆栈实现DFS(Accept)
class Solution {
public:
void solve(vector<vector<char> > &board)
{
int m = board.size();
if (m < 3)
return;
int n = board[0].size();
if (n < 3)
return;
for (int i = 1; i < m - 1; i++)
{
dfs(board, i, 0);
dfs(board, i, n - 1);
}
for (int j = 0; j < n; j++)
{
dfs(board, 0, j);
dfs(board, m - 1, j);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == 'A')
board[i][j] = 'O';
}
}
}
private:
stack<int> st;
void dfs(vector<vector<char> > &board, int x, int y)
{
fill_stack(board, x, y);
while (!st.empty())
{
int curr = st.top();
st.pop();
int i = curr / board[0].size();
int j = curr % board[0].size();
fill_stack(board, i - 1, j);
fill_stack(board, i + 1, j);
fill_stack(board, i, j - 1);
fill_stack(board, i, j + 1);
}
}
void fill_stack(vector<vector<char> > &board, int x, int y)
{
if (x < 0 || x >= board.size() || y<0 || y >= board[0].size() || board[x][y] != 'O')
return;
st.push(x*board[0].size() + y);
board[x][y] = 'A';
}
};
3.BFS
class Solution {
public:
void solve(vector<vector<char> > &board)
{
int m = board.size();
if (m < 3)
return;
int n = board[0].size();
if (n < 3)
return;
for (int i = 1; i < m - 1; i++)
{
bfs(board, i, 0);
bfs(board, i, n - 1);
}
for (int j = 0; j < n; j++)
{
bfs(board, 0, j);
bfs(board, m - 1, j);
}
for (int i = 0; i < m; i++)
{
for (int j = 0; j < n; j++)
{
if (board[i][j] == 'O')
board[i][j] = 'X';
else if (board[i][j] == 'A')
board[i][j] = 'O';
}
}
}
private:
queue<int> Que;
void fill(vector<vector<char> > &board, int x, int y){
if (x < 0 || x >= board.size() || y<0 || y >= board[0].size() || board[x][y] != 'O')
return;
Que.push(x*board[0].size() + y);
board[x][y] = 'A';
}
void bfs(vector<vector<char> > &board,int x, int y){
fill(board, x, y);
while (!Que.empty()){
int curr = Que.front();
Que.pop();
int i = curr / board[0].size();
int j = curr % board[0].size();
fill(board, i - 1, j);
fill(board, i + 1, j);
fill(board, i, j - 1);
fill(board, i, j + 1);
}
}
};