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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.
Example
X X X X
X O O X
X X O X
X O X X
After capture all regions surrounded by ‘X‘, the board should be:
X X X X X X X X X X X X X O X X
public class Solution { /** * @param board a 2D board containing ‘X‘ and ‘O‘ * @return void */ public void surroundedRegions(char[][] board) { // Write your code here if(board == null || board.length <= 1 || board[0] == null || board[0].length <= 1) return; int m = board.length; int n = board[0].length; for(int i = 0; i < m; i++) if(board[i][0] == ‘O‘) flood(board, i, 0); for(int i = 0; i < m; i++) if(board[i][n - 1] == ‘O‘) flood(board, i, n - 1); for(int i = 0; i < n; i++) if(board[0][i] == ‘O‘) flood(board, 0, i); for(int i = 0; i < n; i++) if(board[m - 1][i] == ‘O‘) flood(board, m - 1, i); for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(board[i][j] == ‘O‘) board[i][j] = ‘X‘; } } for(int i = 0; i < m; i++){ for(int j = 0; j < n; j++){ if(board[i][j] == ‘%‘) board[i][j] = ‘O‘; } } return; } public void flood(char[][] board, int i, int j){ int m = board.length; int n = board[0].length; if(i >= m || i < 0 || j >= n || j < 0) return; if(board[i][j] == ‘O‘){ board[i][j] = ‘%‘; flood(board, i + 1, j); flood(board, i - 1, j); flood(board, i, j + 1); flood(board, i, j - 1); } return; } }
lintcode-medium-Surrounded Regions
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原文地址：http://www.cnblogs.com/goblinengineer/p/5362024.html
