# 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.

``````class Solution {
public:
vector<int> gardenNoAdj(int n, vector<vector<int>>& paths) {
vector<int> res(n);
vector<vector<int>> graph(n);
for (auto &path : paths) {
graph[path[0] - 1].push_back(path[1] - 1);
graph[path[1] - 1].push_back(path[0] - 1);
}
for (int i = 0; i < n; ++i) {
vector<bool> colors(5);
for (int j : graph[i]) colors[res[j]] = true;
for (int c = 4; c > 0; --c) {
if (!colors[c]) res[i] = c;
}
}
return res;
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1042