【LeetCode】1042. Flower Planting With No Adjacent 解题报告（Python & C++）
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
题目地址：https://leetcode.com/problems/flower-planting-with-no-adjacent/
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.
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 <= 100000 <= paths.size <= 20000- No garden has 4 or more paths coming into or leaving it.
- It is guaranteed an answer exists.
题目大意
每一个顶点都最多只有3条相邻的边，现在要给每个顶点编号1~4，要求相邻的顶点不能是相同的数字。给出其中任意一种方案。
N表示顶点数，paths表示这两个顶点（编号从1开始）之间有边。
解题方法
图
这个题目背景虽然是花园，但是我相信大家应该都明白了，其实说的是四色定理。
既然是个图论的题目，那么就按照图的方法来解。先构建无向图，对于每个顶点检查其所有相邻顶点的编号，这个顶点用一个没有用过的编号，依次类推。题目也已经说了，解一定存在。
由于每个顶点最多只有三条边，所以时间复杂度是O(N)。
Python代码如下：
class Solution(object):
def gardenNoAdj(self, N, paths):
"""
:type N: int
:type paths: List[List[int]]
:rtype: List[int]
"""
res = [0] * N
graph = [[] for i in range(N)]
for path in paths:
graph[path[0] - 1].append(path[1] - 1)
graph[path[1] - 1].append(path[0] - 1)
for i in range(N):
neighbor_colors = []
for neighbor in graph[i]:
neighbor_colors.append(res[neighbor])
for color in range(1, 5):
if color in neighbor_colors:
continue
res[i] = color
break
return res
C++代码如下：
class Solution {
public:
vector<int> gardenNoAdj(int N, vector<vector<int>>& paths) {
vector<int> res(N, 0);
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) {
unordered_set<int> neighbor_colors;
for (int neighbor : graph[i]) {
neighbor_colors.insert(res[neighbor]);
}
for (int color = 1; color < 5; ++color) {
if (neighbor_colors.count(color))
continue;
res[i] = color;
break;
}
}
return res;
}
};
日期
2019 年 6 月 9 日 —— 简单的题没有难度，需要挑战有难度的才行
