[LeetCode]Surrounded Regions
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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.
For example,
X X X XX O O XX X O XX O X X
After running your function, the board should be:
X X X XX X X XX X X XX O X X
这道题利用深搜会StackOverFlow,思想是所有在最边的'O'及其相连的区域都不会覆盖,找出所有这样的区域,记录在isChecked数组中,遍历修改,其中有一句很重的判重,搞了2个小时才找到。
public class Solution {boolean isChecked[][];int seq = 0; public void solve(char[][] board) { if(board.length==0) return; if(board.length<=2||board[0].length<=2) return; int row = board.length, column = board[0].length; isChecked = new boolean[row][column]; //下面的连个循环用来搜索靠边,且为'O'的部分,然后利用宽搜搜索所有区域 for(int i=0;i<row;i++){ if(!isChecked[i][0]&&board[i][0]=='O') bfs(board,i,0); if(!isChecked[i][column-1]&&board[i][column-1]=='O') bfs(board,i,column-1); } for(int j=1;j<column-1;j++){ if(!isChecked[0][j]&&board[0][j]=='O') bfs(board,0,j); if(!isChecked[row-1][j]&&board[row-1][j]=='O') bfs(board,row-1,j); } //遍历,找到要被修改为'X'的'O' for(int i=1;i<row-1;i++){ for(int j=1;j<column-1;j++){ if(board[i][j]=='O'&&!isChecked[i][j]) board[i][j] = 'X'; } } } private void bfs(char[][] board,int xd,int yd){ int row = board.length, column = board[0].length; Queue<int[]> current = new LinkedList<>(); current.offer(new int[]{xd,yd}); //不记录广搜分层 while(!current.isEmpty()){int[] cd = current.poll();int x=cd[0], y = cd[1];isChecked[x][y] = true;seq++;int move[][] ={{x-1,y},{x,y+1},{x+1,y},{x,y-1}};for(int i=0;i<4;i++){int dx = move[i][0], dy = move[i][1];if(dx<0||dx>=row||dy<0||dy>=column||isChecked[dx][dy]||board[dx][dy]!='O') continue;//下面一句话巨重要,当找到'O'的时候就要记录它已经被搜索过,如果没有记录则其他的路径也会搜索到该点isChecked[dx][dy] = true;current.add(new int[]{dx,dy});} } System.out.println(seq); }}
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