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T21.java
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58 lines (52 loc) · 1.46 KB
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/**
* @author aliyang
* @date 18-6-1 下午4:46
* surrounded-regions：别人解法
* 思路：所有与4条边连接的O都保留，其他都变为X
*/
public class T21 {
private int rowNum=0;
private int colNum=0;
public void solve(char[][] board) {
if(board==null||board.length<=0||board[0].length<=0)
return;
rowNum=board.length;
colNum=board[0].length;
for (int i=0;i<colNum;i++){
dfs(board,0,i);//遍历顶行
dfs(board,rowNum-1,i);//遍历底行
}
for (int i=0;i<rowNum;i++){
dfs(board,i,0);
dfs(board,i,colNum-1);
}
// 将所有的O变为X
for (int i=0;i<rowNum;i++){
for (int j=0;j<colNum;j++){
if (board[i][j]=='O')
board[i][j]='X';
}
}
// 将所有的*变为O
for (int i=0;i<rowNum;i++){
for (int j=0;j<colNum;j++){
if (board[i][j]=='*')
board[i][j]='O';
}
}
}
private void dfs(char[][] board,int row,int col){
if (board[row][col]=='O'){
board[row][col]='*';
// 往上
if (row>1)
dfs(board,row-1,col);
if (col>1)
dfs(board,row,col-1);
if (row<rowNum-1)
dfs(board,row+1,col);
if (col<colNum-1)
dfs(board,row,col+1);
}
}
}