LeetCode 1042. Flower Planting With No Adjacent
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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<Integer, Set<Integer>> graph = new HashMap<>(); 4 for(int i = 0; i < N; i++){ 5 graph.put(i, new HashSet<Integer>()); 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 }
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