LeetCode 1042. Flower Planting With No Adjacent

原题链接在这里：https://leetcode.com/problems/flower-planting-with-no-adjacent/

题目：

You have N gardens, labelled 1 to N.  In each garden, you want to plant one of 4 types of flowers.

paths[i] = [x, y] describes the existence of a bidirectional path from garden x to garden y.

Also, there is no garden that has more than 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]

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]

Note:

• 1 <= N <= 10000
• 0 <= paths.size <= 20000
• No garden has 4 or more paths coming into or leaving it.
• It is guaranteed an answer exists.

题解：

Build the graph with edges. 

For one node, we could check for all its neighbors, what colors have been used. Have a color array to record which color has been used among its neighbors.

Then for this node color, it could choose one color that hasn't been used.

Time Compleixty: O(e + N). e = paths.length.

Space: O(e + N).

AC Java:

 1 class Solution {
 2     public int[] gardenNoAdj(int N, int[][] paths) {
 3         Map> graph = new HashMap<>();
 4         for(int i = 0; i < N; i++){
 5             graph.put(i, new HashSet());
 6         }
 7         
 8         for(int [] p : paths){
 9             int x = p[0] - 1;
10             int y = p[1] - 1;
11             graph.get(x).add(y);
12             graph.get(y).add(x);
13         }
14         
15         int [] res = new int[N];
16         for(int i = 0; i < N; i++){
17             int [] color = new int[5];
18             for(int nei : graph.get(i)){
19                 color[res[nei]] = 1;
20             }
21             
22             for(int c = 4; c >= 1; c--){
23                 if(color[c] != 1){
24                     res[i] = c;
25                 }
26             }
27         }
28         
29         return res;
30     }
31 }