#leetcode#130. Surrounded Regions
阿新 • • 發佈:2018-12-27
iven 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
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思路是flood fill,先把四條邊上的O染色,把相鄰O變成‘1’, 然後再遍歷一般即可
class Solution { public void solve(char[][] board) { if(board == null || board.length < 2 || board[0].length < 2) return; int m = board.length; int n = board[0].length; for(int i = 0; i < n; i++){ fillEdge(board, 0, i); fillEdge(board, m - 1, i); } for(int j = 0; j < m; j++){ fillEdge(board, j, 0); fillEdge(board, j, n - 1); } 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] == '1'){ board[i][j] = 'O'; } } } } private void fillEdge(char[][] board, int row, int col){ if(board[row][col] != 'O'){ return; } int m = board.length; int n = board[0].length; LinkedList<Integer> queue = new LinkedList<>(); queue.offer(row * n + col); while(!queue.isEmpty()){ int idx = queue.poll(); int x = idx / n; int y = idx % n; if(x < 0 || x >= m || y < 0 || y >= n || board[x][y] != 'O'){ continue; } board[x][y] = '1'; queue.offer((x + 1) * n + y); queue.offer((x - 1) * n + y); queue.offer(x * n + y + 1); queue.offer(x * n + y - 1); } } }