114 lines
3.0 KiB
Java
114 lines
3.0 KiB
Java
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//你被给定一个 m × n 的二维网格 rooms ，网格中有以下三种可能的初始化值：
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//
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//
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// -1 表示墙或是障碍物
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// 0 表示一扇门
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// INF 无限表示一个空的房间。然后，我们用 231 - 1 = 2147483647 代表 INF。你可以认为通往门的距离总是小于 2147483647
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//的。
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//
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//
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// 你要给每个空房间位上填上该房间到 最近门的距离 ，如果无法到达门，则填 INF 即可。
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：rooms = [[2147483647,-1,0,2147483647],[2147483647,2147483647,2147483647,-1]
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//,[2147483647,-1,2147483647,-1],[0,-1,2147483647,2147483647]]
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//输出：[[3,-1,0,1],[2,2,1,-1],[1,-1,2,-1],[0,-1,3,4]]
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//
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//
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// 示例 2：
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//
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//
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//输入：rooms = [[-1]]
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//输出：[[-1]]
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//
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//
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// 示例 3：
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//
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//
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//输入：rooms = [[2147483647]]
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//输出：[[2147483647]]
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//
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//
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// 示例 4：
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//
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//
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//输入：rooms = [[0]]
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//输出：[[0]]
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//
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//
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//
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//
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// 提示：
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//
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//
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// m == rooms.length
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// n == rooms[i].length
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// 1 <= m, n <= 250
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// rooms[i][j] 是 -1、0 或 231 - 1
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//
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// Related Topics 广度优先搜索 数组 矩阵
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// 👍 154 👎 0
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package leetcode.editor.cn;
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import java.util.Arrays;
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import java.util.LinkedList;
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import java.util.List;
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import java.util.Queue;
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//286:墙与门
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public class WallsAndGates {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new WallsAndGates().new Solution();
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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private static final int EMPTY = Integer.MAX_VALUE;
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private static final int GATE = 0;
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private final List<int[]> DIRECTIONS = Arrays.asList(
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new int[]{1, 0},
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new int[]{-1, 0},
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new int[]{0, 1},
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new int[]{0, -1}
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);
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public void wallsAndGates(int[][] rooms) {
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int m = rooms.length;
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if (m == 0) {
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return;
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}
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int n = rooms[0].length;
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Queue<int[]> q = new LinkedList<>();
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for (int row = 0; row < m; row++) {
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for (int col = 0; col < n; col++) {
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if (rooms[row][col] == GATE) {
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q.add(new int[]{row, col});
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}
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}
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}
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while (!q.isEmpty()) {
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int[] point = q.poll();
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int row = point[0];
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int col = point[1];
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for (int[] direction : DIRECTIONS) {
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int r = row + direction[0];
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int c = col + direction[1];
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if (r < 0 || c < 0 || r >= m || c >= n || rooms[r][c] != EMPTY) {
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continue;
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}
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rooms[r][c] = rooms[row][col] + 1;
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q.add(new int[]{r, c});
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}
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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