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.
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
Explanation:
Surrounded regions shouldn’t be on the border, which means that any ‘O‘ on the border of the board are not flipped to ‘X‘. Any ‘O‘ that is not on the border and it is not connected to an ‘O‘ on the border will be flipped to ‘X‘. Two cells are connected if they are adjacent cells connected horizontally or vertically.
题目大意：
一个二维数组中含有O和X两个字母，将被X包围的O变成X，边缘上和边缘相连接的O不能变为X。
解法：
1.将边缘和边缘相连接的O变为#,采用dfs
2.将数组中的O变为X
3.将数组中的#变为O
例如;
X X X X X X X X X X X X
X X O X -> X X O X -> X X X X
X O X X X # X X X O X X
X O X X X # X X X O X XJava：
class Solution {
void dfs(char[][] board,int i,int j){
board[i][j]=‘#‘;
int m=board.length,n=board[0].length;
if(i+1<m && board[i+1][j]==‘O‘) dfs(board,i+1,j);
if(j+1<n && board[i][j+1]==‘O‘) dfs(board,i,j+1);
if(i-1>=0 && board[i-1][j]==‘O‘) dfs(board,i-1,j);
if(j-1>=0 && board[i][j-1]==‘O‘) dfs(board,i,j-1);
}
public void solve(char[][] board) {
if(board.length==0 || board[0].length==0) return;
int m=board.length,n=board[0].length;
for(int i=0;i<n;i++){
if(board[0][i]==‘O‘) dfs(board,0,i);
if(board[m-1][i]==‘O‘) dfs(board,m-1,i);
}
for(int i=0;i<m;i++){
if(board[i][0]==‘O‘) dfs(board,i,0);
if(board[i][n-1]==‘O‘) dfs(board,i,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‘;
}
}
for(int i=0;i<m;i++){
for(int j=0;j<n;j++){
if(board[i][j]==‘#‘) board[i][j]=‘O‘;
}
}
}
}
leetcode [130]Surrounded Regions
原文：https://www.cnblogs.com/xiaobaituyun/p/10715812.html