我觉得这道题和传统的用动规或者贪心等算法的题目不同。按照题目的意思，就是将被‘X’围绕的‘O’区域找出来，然后覆盖成‘X’。
那问题就变成两个子问题：
1. 找到‘O’区域，可能有多个区域，每个区域‘O’都是相连的；
2. 判断‘O’区域是否是被‘X’包围。
我采用树的宽度遍历的方法，找到每一个‘O’区域，并为每个区域设置一个value值，为0或者1,1表示是被‘X’包围，0则表示不是。是否被‘X’包围就是看‘O’区域的边界是否是在2D数组的边界上。
下面是具体的AC代码:
class BoardNode{
int x;
int y;
public BoardNode(int x, int y){
this.x = x;
this.y = y;
}
}
1 /**
2 * Given a 2D board containing ‘X‘ and ‘O‘, capture all regions surrounded by ‘X‘.
3 * A region is captured by flipping all ‘O‘s into ‘X‘s in that surrounded region.
4 * @param board
5 */
6 public void solve(char[][] board){
7 //empty 2D board
8 if(board.length == 0)
9 return;
10 //if value == 1,means the group is surrended by ‘X‘, if value == 0 , means it is not
11 HashMap<LinkedList<BoardNode>, Integer> groups = new HashMap<LinkedList<BoardNode>, Integer>();
12 int xB = board.length;
13 int yB = board[0].length;
14 if(xB == 1 || yB == 1)
15 return;
16 boolean[][] flag = new boolean[xB][yB];
17 for(int i=0;i<xB;i++)
18 for(int j=0;j<yB;j++)
19 flag[i][j] = true;
20 for(int i=0;i<xB;i++)
21 {
22 for(int j=0;j<yB;j++){
23 if(board[i][j]==‘O‘ && flag[i][j])
24 {
25 //a new group
26 LinkedList<BoardNode> tree = new LinkedList<BoardNode>();
27 //the corresponding value of the group
28 int value = 1;
29 if(i==0 || i==xB-1 || j==0 || j==yB-1)
30 value = 0;
31 //bread first search for the group
32 tree.offer(new BoardNode(i,j));
33 flag[i][j] = false;
34 int point = 0;
35 while(point<tree.size()){
36 BoardNode temp = tree.get(point++);
37 //left neighbor
38 if(temp.y+1==yB-1 && board[temp.x][temp.y+1]== ‘O‘)
39 value = 0;
40 if(temp.y+1<yB && flag[temp.x][temp.y+1] && board[temp.x][temp.y+1]== ‘O‘)
41 {
42 flag[temp.x][temp.y+1] = false;
43 tree.offer(new BoardNode(temp.x,temp.y+1));
44 }
45 //down neighbor
46 if(temp.x+1 == xB-1 && board[temp.x+1][temp.y] == ‘O‘)
47 value = 0;
48 if(temp.x+1<xB && flag[temp.x+1][temp.y] && board[temp.x+1][temp.y] == ‘O‘)
49 {
50 flag[temp.x+1][temp.y]=false;
51 tree.offer(new BoardNode(temp.x+1,temp.y));
52 }
53 //up neighbor
54 if(temp.x-1 == 0 && board[temp.x-1][temp.y]==‘O‘)
55 value = 0;
56 if(temp.x-1>=0 && flag[temp.x-1][temp.y] && board[temp.x-1][temp.y] == ‘O‘){
57 flag[temp.x-1][temp.y]=false;
58 tree.offer(new BoardNode(temp.x-1,temp.y));
59 }
60 //left neighbor
61 if(temp.y-1==0 && board[temp.x][temp.y-1]==‘O‘)
62 value = 0;
63 if(temp.y-1>=0 && flag[temp.x][temp.y-1] && board[temp.x][temp.y-1]==‘O‘)
64 {
65 flag[temp.x][temp.y-1] = false;
66 tree.offer(new BoardNode(temp.x,temp.y-1));
67 }
68 }
69 groups.put(tree, value);
70 }
71 }
72 }
73 //flipping the region surrounded by ‘X‘
74 for(Entry<LinkedList<BoardNode>,Integer> e: groups.entrySet())
75 {
76 if(e.getValue() == 1)
77 {
78 for(BoardNode bn : e.getKey())
79 board[bn.x][bn.y] = ‘X‘;
80 }
81 }
82 }
LeetCode OJ - Surrounded Regions,布布扣,bubuko.com
时间： 2024-10-03 01:13:23