130. Surrounded Regions
题目
Given a 2D board containing 'X' and 'O' (the letter 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
解析
- 核心思想：只有边界上’O’的位置组成的片区不会被’X’包围。因此先对边界上的’O’遍历之后暂存为’‘。非’‘的’O’即被’X’包围了。
此题考查DFS/BFS的思想，就像走迷宫问题，入口和出口位置不同，DFS/BFS的效率不一样，递归式的DFS 会出现stackoverflow 或者run time的问题
bug :发现leetcode系统必须检查输入为空的情况！否则出现
"Runtime Error Message:reference binding to null pointer of type 'struct value_type' Last executed input: []之类的错误！！！
class Solution {
public:
void bfs(vector<vector<char>> &board,int cur_i,int cur_j,int row,int col)
{
if(board[cur_i][cur_j]!='O')
return;
board[cur_i][cur_j]='*';
queue<pair<int, int >> q;
q.push(make_pair(cur_i,cur_j));
while(!q.empty())
{
int i=q.front().first;
int j=q.front().second;
q.pop();
//board[i][j]='*';
if(i-1>=0&&board[i-1][j]=='O')
{
board[i-1][j]='*';
q.push(make_pair(i-1,j));
}
if(i+1<row&& board[i+1][j]=='O') //i+1<=row bug:vector边界溢出，花了很长时间才找到！！！
{
board[i+1][j]='*';
q.push(make_pair(i+1,j));
}
if(j-1>=0&&board[i][j-1]=='O')
{
board[i][j-1]='*';
q.push(make_pair(i,j-1));
}
if(j+1<col&&board[i][j+1]=='O') //j+1<=col
{
board[i][j+1]='*';
q.push(make_pair(i,j+1));
}
}
}
void solve(vector<vector<char>> &board) {
//bfs
if (board.empty()) //bug :发现leetcode系统必须检查输入为空的情况！
return;
int row=board.size();
int col=board[0].size();
if(row==0||col==0)
return;
for(int i=0;i<row;i++)
{
if(board[i][0]=='O')
bfs(board,i,0,row,col);
if(board[i][col-1]=='O')
bfs(board,i,col-1,row,col);
}
for(int j=0;j<col;j++)
{
if(board[0][j]=='O')
bfs(board,0,j,row,col);
if(board[row-1][j]=='O')
bfs(board,row-1,j,row,col);
}
for(int i=0;i<row;i++)
for(int j=0;j<col;j++)
{
board[i][j]= (board[i][j]=='*')?'O':'X';
}
}
};
- 这样考虑时间复杂度
for(int i=0;i<row;i++)
for(int j=0;j<col;j++) //直接也有复杂度的问题
{
if(board[i][j]=='O')
{
if(i==0||i==row-1||j==0||j==col-1)
{
bfs(board,i,j,row,col); //广度优先搜索使用队列
}
}
}