题目链接:130. 被围绕的区域
思路:使用两遍dfs,第一遍找到可以被替换区域的可进入点并记录,第二遍就从所有的可进入点入手遍历区域内所有点并替换。
这是我的思路,感觉还是挺新颖的(应该很少有人这样想我觉得),算是抛砖引玉了。
class Solution {
int[] dx = {0, 0, 1, -1};
int[] dy = {1, -1, 0, 0};
boolean exit = false;// 判断是否可以退出
public void solve(char[][] board) {
// 定义一个记录矩阵判断某个点是否被走过
int[][] record = new int[board.length][board[0].length];
// 用一个数列存所有的可以进入点
ArrayList<int[]> list = new ArrayList<>();
// 遍历这个矩阵
for(int i = 0; i < board.length; i++) {
for (int j = 0; j < board[0].length; j++) {
// 如果这个点没有被走过,并且是'O',则开始深度优先搜索
if (record[i][j] == 0 && board[i][j] == 'O') {// 某个区域的进入点
record[i][j] = 1;
exit = false;
dfsOnlyWalk(board, record, i, j);
if(!exit) {
list.add(new int[]{i, j});
}
}
}
}
for(int[] ints : list) {
int x = ints[0], y = ints[1];
board[x][y] = 'X';
dfsOnlyMark(board, record, x, y);
}
}
public void dfsOnlyWalk(char[][] board, int[][] record, int i, int j) {
//区域中有点位于边缘
if(i == 0 || i == board.length - 1 || j == 0 || j == board[0].length - 1) {
exit = true;
}
for(int k = 0; k < 4; k++) {
int xx = i + dx[k];
int yy = j + dy[k];
if(xx >= 0 && xx < board.length && yy >= 0 && yy < board[0].length && record[xx][yy] == 0 && board[xx][yy] == 'O') {
record[xx][yy] = 1;
dfsOnlyWalk(board, record, xx, yy);
}
}
}
public void dfsOnlyMark(char[][] board, int[][] record, int i, int j) {
for(int k = 0; k < 4; k++) {
int xx = i + dx[k];
int yy = j + dy[k];
if(xx >= 0 && xx < board.length && yy >= 0 && yy < board[0].length && board[xx][yy] == 'O') {
board[xx][yy] = 'X';
dfsOnlyMark(board, record, xx, yy);
}
}
}
public static void main(String[] args) {
Solution solution = new Solution();
/*
* [["O","X","O","O","X","X"],["O","X","X","X","O","X"],["X","O","O","X","O","O"],["X","O","X","X","X","X"],["O","O","X","O","X","X"],["X","X","O","O","O","O"]]
* */
char[][] board = {{'O','X','O','O','X','X'},{'O','X','X','X','O','X'},{'X','O','O','X','O','O'},{'X','O','X','X','X','X'},{'O','O','X','O','X','X'},{'X','X','O','O','O','O'}};
solution.solve(board);
for(char[] chars : board) {
for(char c : chars) {
System.out.print(c + " ");
}
System.out.println();
}
}
}
736

被折叠的 条评论
为什么被折叠?



