LeetCode-130-Surrounded Regions
Posted 无名路人甲
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算法描述:
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.
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
Explanation:
Surrounded regions shouldn’t be on the border, which means that any ‘O‘
on the border of the board are not flipped to ‘X‘
. Any ‘O‘
that is not on the border and it is not connected to an ‘O‘
on the border will be flipped to ‘X‘
. Two cells are connected if they are adjacent cells connected horizontally or vertically.
解题思路:深度优先搜索或者广度优先搜索。从边界开始搜索,将符合条件的值得索引放入队列中,并用额外的矩阵标记访符合要求的数值位置。
void solve(vector<vector<char>>& board) { int m = board.size(); if(m==0) return; int n = board[0].size(); if(n==0) return; vector<vector<int>> alive(m, vector<int>(n, 0)); queue<pair<int, int>> que; for(int i=0; i< m; i++){ if(board[i][0] == ‘O‘){ alive[i][0] =1; que.push(make_pair(i,0)); } if(board[i][n-1] == ‘O‘){ alive[i][n-1] =1; que.push(make_pair(i,n-1)); } } for(int i=0; i< n; i++){ if(board[0][i] == ‘O‘){ alive[0][i] =1; que.push(make_pair(0,i)); } if(board[m-1][i] == ‘O‘){ alive[m-1][i] =1; que.push(make_pair(m-1,i)); } } while(!que.empty()){ pair<int,int> temp = que.front(); que.pop(); int dx[] = {-1,0,1,0}; int dy[] = {0,1,0,-1}; for(int i = 0; i < 4; i++){ int ax = temp.first + dx[i]; int ay = temp.second + dy[i]; if(ax >= 0 && ay >=0 && ax < m && ay < n && alive[ax][ay] ==0 && board[ax][ay] == ‘O‘){ alive[ax][ay] = 1; que.push(make_pair(ax,ay)); } } } for(int i=0; i < m; i++){ for(int j=0; j <n; j++){ if(alive[i][j]==0 && board[i][j]==‘O‘) board[i][j] =‘X‘; } } }
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