【Surrounded Regions】cpp
题目:
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
代码:
class Solution {
public:
void solve(vector<vector<char>>& board) {
const int ROW = board.size();
if ( ROW< ) return;
const int COL = board[].size();
// first row
for ( int i = ; i < COL; ++i )
{if ( board[][i]=='O') { board[][i] = 'M'; Solution::bfs(board, , i, ROW, COL); }}
// last row
for ( int i = ; i < COL; ++i )
{if ( board[ROW-][i]=='O' ) { board[ROW-][i] = 'M'; Solution::bfs(board, ROW-, i, ROW, COL); }}
// first col
for ( int i = ; i < ROW; ++i )
{if ( board[i][]=='O' ) { board[i][] = 'M'; Solution::bfs(board, i, , ROW, COL); }}
// last col
for ( int i = ; i < ROW; ++i )
{if ( board[i][COL-]=='O' ) { board[i][COL-]='M'; Solution::bfs(board, i, COL-, ROW, COL); }}
// flipping surrounded regions
for ( int i = ; i < ROW; ++i ){
for ( int j = ; j < COL; ++j ){
if ( board[i][j]=='O' ) { board[i][j]='X'; continue; }
if ( board[i][j]=='M' ) { board[i][j]='O'; continue; }
}
}
}
static void bfs(vector<vector<char> >& board, int row, int col, int ROW, int COL )
{
queue<pair<int, int> > que;
que.push(make_pair(row, col));
while ( !que.empty() ){
int r = que.front().first, c = que.front().second;
que.pop();
// up left down right
if ( r-> && board[r-][c]=='O' ) { board[r-][c]='M'; que.push(make_pair(r-, c)); }
if ( c-> && board[r][c-]=='O') { board[r][c-]='M'; que.push(make_pair(r, c-)); }
if ( r+<ROW- && board[r+][c]=='O' ) { board[r+][c]='M'; que.push(make_pair(r+, c)); }
if ( c+<COL- && board[r][c+]=='O' ) { board[r][c+]='M'; que.push(make_pair(r, c+)); }
}
}
};
tips:
参考的BFS的思路(http://yucoding.blogspot.sg/2013/08/leetcode-question-131-surrounded-regions.html)
先说大体思路
这个思路正好与常规的思路逆过来:
1. 常规的思路是从wall里面挨个BFS遍历,如果从一个'O'出发的所有点都满足封住了,则这BFS走过的点都被封住了,最后赋值为'X'(这样会超时)
2. 而这道题的思路比较好的是逆过来,从边界的'O'出发BFS(因为只要跟边界的'O'连上了,就必然封不住了;而没有与边界的'O'边界连上的'O'自然被封住了)
再说实现细节:
1. 为了不使用额外空间,对于由边界'O'经BFS得到的'O'都置为'M'(M有两个意思:一是访问过了,不用再访问了;二是这个点再最后要恢复为'O')
2. BFS的时候,终止条件第一次直接写的是que.empty(),一直不对;后来改成了!que.empty()才对,这个低级错误不要再犯。
=========================================
第二次过这道题,有个细节没有注意:
如果发现一个点四周的点是要保留的‘O’时,一定要马上将其设置为‘M’,然后再进行BFS;如果不马上设置为‘M',那么在BFS的过程中,可能会重复过这个点。
class Solution {
public:
void solve(vector<vector<char> >& board)
{
if ( board.size()< ) return;
const int m = board.size();
const int n = board[].size();
// search from first row
for ( int i=; i<n; ++i )
{
if ( board[][i]=='O' ) Solution::bfs(board, , i, m, n);
}
// search from last row
for ( int i=; i<n; ++i )
{
if ( board[m-][i]=='O' ) Solution::bfs(board, m-, i, m, n);
}
// search from first column
for ( int i=; i<m; ++i )
{
if ( board[i][]=='O' ) Solution::bfs(board, i, , m, n);
}
// search from last column
for ( int i=; i<m; ++i )
{
if ( board[i][n-]=='O' ) Solution::bfs(board, i, n-, m, n);
}
// trans 'O' to 'X' & trans 'M' to 'O'
for ( int i=; i<m; ++i )
{
for ( int j=; j<n; ++j )
{
if ( board[i][j]=='O' )
{
board[i][j]='X';
continue;
}
if ( board[i][j]=='M' )
{
board[i][j]='O';
}
}
}
}
static void bfs(vector<vector<char> >& board, int r, int c, int ROW, int COL)
{
board[r][c] = 'M';
queue<pair<int, int> > curr;
queue<pair<int, int> > next;
curr.push(make_pair(r, c));
while ( !curr.empty() )
{
while ( !curr.empty() )
{
int i = curr.front().first;
int j = curr.front().second;
curr.pop();
// up
if ( i->= && board[i-][j]=='O' )
{
board[i-][j] = 'M';
next.push(make_pair(i-, j));
}
// down
if ( i+<ROW && board[i+][j]=='O' )
{
board[i+][j] = 'M';
next.push(make_pair(i+, j));
}
// left
if ( j->= && board[i][j-]=='O' )
{
board[i][j-] = 'M';
next.push(make_pair(i, j-));
}
// right
if ( j+<COL && board[i][j+]=='O' )
{
board[i][j+] = 'M';
next.push(make_pair(i, j+));
}
}
swap(next, curr);
}
}
};
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