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[LeetCode] 1042. Flower Planting With No Adjacent

发布时间 2023-04-16 02:19:40作者: CNoodle

You have n gardens, labeled from 1 to n, and an array paths where paths[i] = [xi, yi] describes a bidirectional path between garden xi to garden yi. In each garden, you want to plant one of 4 types of flowers.

All gardens have at most 3 paths coming into or leaving it.

Your task is to choose a flower type for each garden such that, for any two gardens connected by a path, they have different types of flowers.

Return any such a choice as an array answer, where answer[i] is the type of flower planted in the (i+1)th garden. The flower types are denoted 1, 2, 3, or 4. It is guaranteed an answer exists.

Example 1:

Input: n = 3, paths = [[1,2],[2,3],[3,1]]
Output: [1,2,3]
Explanation:
Gardens 1 and 2 have different types.
Gardens 2 and 3 have different types.
Gardens 3 and 1 have different types.
Hence, [1,2,3] is a valid answer. Other valid answers include [1,2,4], [1,4,2], and [3,2,1].

Example 2:

Input: n = 4, paths = [[1,2],[3,4]]
Output: [1,2,1,2]

Example 3:

Input: n = 4, paths = [[1,2],[2,3],[3,4],[4,1],[1,3],[2,4]]
Output: [1,2,3,4]

Constraints:

  • 1 <= n <= 104
  • 0 <= paths.length <= 2 * 104
  • paths[i].length == 2
  • 1 <= xi, yi <= n
  • xi != yi
  • Every garden has at most 3 paths coming into or leaving it.

不邻接植花。

有 n 个花园，按从 1 到 n 标记。另有数组 paths ，其中 paths[i] = [xi, yi] 描述了花园 xi 到花园 yi 的双向路径。在每个花园中，你打算种下四种花之一。

另外，所有花园 最多 有 3 条路径可以进入或离开.

你需要为每个花园选择一种花，使得通过路径相连的任何两个花园中的花的种类互不相同。

以数组形式返回 任一 可行的方案作为答案 answer，其中 answer[i] 为在第 (i+1) 个花园中种植的花的种类。花的种类用  1、2、3、4 表示。保证存在答案。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/flower-planting-with-no-adjacent
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

思路是 DFS + 染色。题目说每个花园你都要种花，那么意味着你需要遍历整个图。这里我用 hashmap 把图建立起来，之后开始遍历，遍历的时候我还用了一个长度为 5 的 boolean 数组记录每个 node 的邻居节点是否有被染色过。长度为什么是 5 是因为题目说可以用1 - 4来染色，为了处理 0 方便我就把长度定为 5。注意题目给的 path 里面的数字在建图的时候需要 - 1。其余部分参见代码注释。

时间O(V + E)

空间O(n)

Java实现

 1 class Solution {
 2     public int[] gardenNoAdj(int n, int[][] paths) {
 3         // 建图
 4         HashMap<Integer, Set<Integer>> g = new HashMap<>();
 5         for (int i = 0; i < n; i++) {
 6             g.put(i, new HashSet<>());
 7         }
 8         for (int[] p : paths) {
 9             int from = p[0] - 1;
10             int to = p[1] - 1;
11             g.get(from).add(to);
12             g.get(to).add(from);
13         }
14 
15         // 记录每个位置填什么颜色
16         int[] res = new int[n];
17         for (int i = 0; i < n; i++) {
18             // 如果邻居节点有上过色的，标记一下
19             boolean[] used = new boolean[5];
20             for (int j : g.get(i)) {
21                 used[res[j]] = true;
22             }
23             for (int c = 1; c <= 4; c++) {
24                 // 给当前位置上色
25                 if (used[c] == false) {
26                     res[i] = c;
27                 }
28             }
29         }
30         return res;
31     }
32 }

 

LeetCode 题目总结