标签：style bsp boa queue capture lis boolean continue ret
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
像围棋一样，去掉被包围的O，刚开始用了HashSet<Integer>但是内存溢出了。
public class Solution { public void solve(char[][] board) { int row = board.length; if( row < 2 ) return ; int col = board[0].length; if( col < 2 ) return ; for( int i = 0;i<row;i++){ for( int j = 0;j<col;j++){ if( board[i][j] == ‘O‘){ HashSet<Integer> set = new HashSet<Integer>(); if( helper(board,i,j,set,false) ){ helpset(‘o‘,set,board); }else helpset(‘X‘,set,board); } } } for( int i = 0;i<row;i++){ for( int j = 0;j< col;j++){ if( board[i][j] == ‘o‘) board[i][j] = ‘O‘; } } } public void helpset(char ch ,HashSet<Integer> set ,char[][] board){ for( int i : set ){ int row = i/board[0].length; int col = i%board[0].length; board[row][col] = ch; } } public boolean helper(char[][] board,int num1, int num2,HashSet<Integer> set,boolean flag){ boolean result = flag; if( num1-1 >= 0 && board[num1-1][num2] == ‘O‘){ int num = (num1-1)*board[0].length+num2; if( !set.contains(num) ){ set.add(num); result = result || helper(board,num1-1,num2,set,flag); } } if( num1+1 < board.length && board[num1+1][num2] == ‘O‘){ int num = (num1+1)*board[0].length+num2; if( !set.contains(num) ){ set.add(num); result = result || helper(board,num1+1,num2,set,flag); } } if( num2-1 >= 0 && board[num1][num2-1] == ‘O‘){ int num = num1*board[0].length+num2-1; if( !set.contains(num) ){ set.add(num); result = result || helper(board,num1,num2-1,set,flag); } } if( num2+1 < board[0].length && board[num1][num2+1] == ‘O‘){ int num = num1*board[0].length+num2+1; if( !set.contains(num) ){ set.add(num); result = result || helper(board,num1,num2+1,set,flag); } } if( num1 == 0 || num1 == board.length-1 || num2 == 0 || num2 == board[0].length-1) return true; return result; } }
2、换用思路，先把所有的O放入一个set中，然后在这个set中进行判断，虽然ac但是还是耗时还是较长。
public class Solution { public void solve(char[][] board) { int row = board.length; if( row < 2 ) return ; int col = board[0].length; if( col < 2 ) return ; HashSet<Integer> set = new HashSet<Integer>(); for( int i = 0;i<row;i++){ for( int j = 0;j<col;j++){ if( board[i][j] == ‘O‘) set.add(i*board[0].length+j); } } while( !set.isEmpty() ){ helper(set,col,row,board); } return ; } public void helper(HashSet<Integer> set,int len,int row,char[][] board){ boolean result = false; Queue<Integer> queue = new LinkedList<Integer>(); int[] set2 = new int[set.size()]; int num = set.iterator().next(); set2[0] = num; int i = 1; queue.add(num); set.remove(num); if (num % len == 0 || num % len == len - 1 || num / len == 0 || num / len == row - 1) result = true; while( !queue.isEmpty() ){ num = (Integer) queue.poll(); if( set.contains(num+1) ){ set2[i] = num+1; i++; queue.add(num+1); set.remove(num+1); if ((num+1) % len == 0 || (num+1) % len == len - 1 || (num+1) / len == 0 || (num+1) / len == row - 1) result = true; } if( set.contains(num-1)){ set2[i] = (num-1); i++; queue.add(num-1); set.remove(num-1); if ((num-1) % len == 0 ||( num-1) % len == len - 1 || (num-1) / len == 0 || (num-1) / len == row - 1) result = true; } if( set.contains(num+len) ){ set2[i] = (num+len); i++; queue.add(num+len); set.remove(num+len); if ((num+len) % len == 0 || (num+len )% len == len - 1 || (num+len) / len == 0 || (num+len) / len == row - 1) result = true; } if( set.contains(num-len) ){ set2[i] = (num-len); i++; queue.add(num-len); set.remove(num-len); if ((num-len) % len == 0 || (num-len) % len == len - 1 || (num-len) / len == 0 || (num-len) / len == row - 1) result = true; } } if( result == true ) return ; else helpset(‘X‘,set2,i,board); } public void helpset(char ch ,int[] set,int num ,char[][] board){ for( int i = 0;i<num;i++){ int row = set[i]/board[0].length; int col = set[i]%board[0].length; board[row][col] = ch; } } }
3、使用BFS和队列。
public class Solution { public void solve(char[][] board) { int row = board.length; if( row < 2 ) return ; int col = board[0].length; if( col < 2 ) return ; Queue<Integer> queue = new LinkedList<Integer>(); for( int i = 0;i<col;i++){ if( board[0][i] == ‘O‘ ) queue.add(i); if( board[row-1][i] == ‘O‘) queue.add((row-1)*col+i); } for( int i = 0;i < row ;i++){ if( board[i][col-1] == ‘O‘) queue.add(i*col+col-1); if( board[i][0] == ‘O‘) queue.add(i*col); } while( !queue.isEmpty() ){ int num = queue.poll(); int x = num/col,y = num%col; if( board[x][y] != ‘O‘) continue; board[x][y] = ‘o‘; if( x-1>=0 && board[x-1][y] == ‘O‘) queue.add(num-col); if( x+1<row && board[x+1][y] == ‘O‘) queue.add(num+col); if( y-1>=0 && board[x][y-1] == ‘O‘) queue.add(num-1); if( y+1<col && board[x][y+1] == ‘O‘) queue.add(num+1); } for( int i = 0;i<row;i++){ for( int j = 0;j<col;j++){ if( board[i][j] == ‘O‘) board[i][j] = ‘X‘; else if( board[i][j] == ‘o‘) board[i][j] = ‘O‘; } } return ; } }
4、不使用队列，直接用BFS。
public class Solution { public void solve(char[][] board) { if (board.length == 0) return; int row = board.length; int col = board[0].length; for (int i = 0; i < row; i++) { if (board[i][0] == ‘O‘) dfs(board, i, 0); if (board[i][col-1] == ‘O‘) dfs(board, i, col-1); } for (int i = 1; i < col - 1; i++) { if (board[0][i] == ‘O‘) dfs(board, 0, i); if (board[row - 1][i] == ‘O‘) dfs(board,row-1, i); } for (int i = 0; i < board.length; i++) { for (int j = 0; j < board[0].length; j++) { if (board[i][j] == ‘1‘) { board[i][j] = ‘O‘; } else { board[i][j] = ‘X‘; } } } return; } public void dfs(char[][] board, int m, int n) { if (board[m][n] != ‘O‘) return; board[m][n] = ‘1‘; if (m < board.length - 2) dfs(board, m + 1, n); if (m > 1) dfs(board, m - 1, n); if (n < board[0].length - 2) dfs(board, m, n + 1); if (n > 1) dfs(board, m, n - 1); } }
leetcode 130. Surrounded Regions----- java
标签：style bsp boa queue capture lis boolean continue ret
原文地址：http://www.cnblogs.com/xiaoba1203/p/6037045.html
