题目描述

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.

样例

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]

算法1

(贪心) $O(n^2)$

建立无向图的连接，遍历每一个节点，用color数组标记该节点的邻节点j颜色res[j]为已分配，然后在四种颜色中找到一个未分配的颜色给i节点，if color[c]==0, res[i]=c

时间复杂度

参考文献

C++ 代码

class Solution {
public:
    vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
        vector<int> res(N);
        vector<vector<int>> G(N);
        for (vector<int>& p : paths) {
            G[p[0] - 1].push_back(p[1] - 1);
            G[p[1] - 1].push_back(p[0] - 1);
        }
        for (int i = 0; i < N; ++i) {
            int colors[5] = {};
            for (int j : G[i])           //mark neighbour as used
                colors[res[j]] = 1;
            for (int c = 1; c <= 4; ++c) 
            {
                if (!colors[c])          //assign a unused color to i node
                {
                    res[i] = c;
                    break; 
                }
            }
        }
        return res;
    }

};