[LeetCode] 1042. Flower Planting With No Adjacent

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
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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

class Solution {
    public int[] gardenNoAdj(int n, int[][] paths) {
        // 建图
        HashMap<Integer, Set<Integer>> g = new HashMap<>();
        for (int i = 0; i < n; i++) {
            g.put(i, new HashSet<>());
        }
        for (int[] p : paths) {
            int from = p[0] - 1;
            int to = p[1] - 1;
            g.get(from).add(to);
            g.get(to).add(from);
        }

        // 记录每个位置填什么颜色
        int[] res = new int[n];
        for (int i = 0; i < n; i++) {
            // 如果邻居节点有上过色的，标记一下
            boolean[] used = new boolean[5];
            for (int j : g.get(i)) {
                used[res[j]] = true;
            }
            for (int c = 1; c <= 4; c++) {
                // 给当前位置上色
                if (used[c] == false) {
                    res[i] = c;
                }
            }
        }
        return res;
    }
}

 

LeetCode 题目总结