231003 거리두기 확인하기

Jongleee·2023년 10월 3일
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TIL

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private static final int[] dx = { 0, 1, 0, -1 };
private static final int[] dy = { -1, 0, 1, 0 };

public int[] solution(String[][] places) {
	int[] result = new int[places.length];
	for (int i = 0; i < places.length; i++) {
		result[i] = isCorrect(places[i]);
	}
	return result;
}

private int isCorrect(String[] board) {
	for (int i = 0; i < board.length; i++) {
		for (int j = 0; j < board[i].length(); j++) {
			if (board[i].charAt(j) == 'P') {
				if (!isSafe(board, i, j))
					return 0;
			}
		}
	}
	return 1;
}

private boolean isSafe(String[] p, int x, int y) {
	Queue<int[]> queue = new LinkedList<>();
	queue.add(new int[] { x, y });
	boolean[][] visited = new boolean[p.length][p[0].length()];

	while (!queue.isEmpty()) {
		int[] temp = queue.poll();
		int currX = temp[0];
		int currY = temp[1];

		visited[currX][currY] = true;

		for (int i = 0; i < 4; i++) {
			int nx = currX + dx[i];
			int ny = currY + dy[i];

			if (!isValid(p, nx, ny) || visited[nx][ny]) {
				continue;
			}

			int manhattanDistance = Math.abs(x - nx) + Math.abs(y - ny);

			if (p[nx].charAt(ny) == 'P' && manhattanDistance <= 2) {
				return false;
			} else if (p[nx].charAt(ny) == 'O' && manhattanDistance < 2) {
				queue.add(new int[] { nx, ny });
			}
		}
	}
	return true;
}

private boolean isValid(String[] p, int x, int y) {
	return x >= 0 && y >= 0 && x < p.length && y < p[0].length();
}

출처:https://school.programmers.co.kr/learn/courses/30/lessons/81302

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