Middle-题目85:310. Minimum Height Trees
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题目原文:
For a undirected graph with tree characteristics, we can choose any node as the root. The result graph is then a rooted tree. Among all possible rooted trees, those with minimum height are called minimum height trees (MHTs). Given such a graph, write a function to find all the MHTs and return a list of their root labels.
Format
The graph contains n nodes which are labeled from 0 to n - 1. You will be given the number n and a list of undirected edges (each edge is a pair of labels).
You can assume that no duplicate edges will appear in edges. Since all edges are undirected, [0, 1] is the same as [1, 0] and thus will not appear together in edges.
Example 1:
Given n = 4, edges = [[1, 0], [1, 2], [1, 3]]
0 | 1 / \ 2 3
return [1]
Example 2:
Given n = 6, edges = [[0, 3], [1, 3], [2, 3], [4, 3], [5, 4]]
0 1 2 \ | / 3 | 4 | 5
return [3, 4]
题目大意:
给出一个树,它的任何一个顶点都可以看做树的根节点,写一个函数,输出使得树高度最小的顶点的列表。
输入参数是顶点个数和边的列表(用二维数组表示)。
例如,输入n = 4, 边列表= [[1, 0], [1, 2], [1, 3]]
则以1为根节点的树的高度最小。
题目分析:
不管树有多复杂,使得高度最小的顶点至多有2个。(我也想不出证明过程……)
暴力解法是对每个节点求高度。这里有一个比较好的算法,即每轮去掉度为1的节点,直到顶点剩下1个或2个(显然不可能是3个或更多,因为没有环),那么这1个或2个节点即为所求。
源码:(language:java)
public class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { Map<Integer, Set<Integer>> adjList = new HashMap<Integer,Set<Integer>>(); boolean[] isLeaf = new boolean[n]; int[] degrees = new int[n]; int nextVertical = 0, currentVerticals = n; for(int i = 0; i<n;i++) adjList.put(i, new HashSet<Integer>()); for(int[] edge : edges) { adjList.get(edge[0]).add(edge[1]); degrees[edge[0]]++; adjList.get(edge[1]).add(edge[0]); degrees[edge[1]]++; } while(currentVerticals > 2) { for(int i = 0;i<n;i++) { if(degrees[i]==1) { isLeaf[i] = true; } } for(int i = 0;i<n;i++) { if(isLeaf[i]) { nextVertical = adjList.get(i).iterator().next(); degrees[i]--; degrees[nextVertical]--; adjList.get(i).remove(nextVertical); if(degrees[i]==0) currentVerticals--; adjList.get(nextVertical).remove(i); if(degrees[nextVertical]==0) currentVerticals--; isLeaf[i] = false; } } } List<Integer> list = new ArrayList<Integer>(); for(Integer key : adjList.keySet()) { if(adjList.get(key).size()!=0) list.add(key); } if(n==1) list.add(0); if(currentVerticals==0) list.add(nextVertical); return list; }}
这里给出discuss中一位大神给出的算法,思路是一样的,但省略了很多循环:
public class Solution { public List<Integer> findMinHeightTrees(int n, int[][] edges) { if (n == 1) return Collections.singletonList(0); List<Set<Integer>> adj = new ArrayList<>(n); for (int i = 0; i < n; i++) adj.add(new HashSet<>()); for (int[] edge : edges) { adj.get(edge[0]).add(edge[1]); adj.get(edge[1]).add(edge[0]); } List<Integer> leaves = new ArrayList<>(); for (int i = 0; i < n; i++) if (adj.get(i).size() == 1) leaves.add(i); while (n > 2) { n -= leaves.size(); List<Integer> newLeaves = new ArrayList<>(); for (int i : leaves) { int j = adj.get(i).iterator().next(); adj.get(j).remove(i); if (adj.get(j).size() == 1) newLeaves.add(j); } leaves = newLeaves; } return leaves; }}
成绩:
算法一(我写的):170ms,beats 14.07%,众数56ms,5.12%
算法二(from discuss):58ms,beats 68.02%
Cmershen的碎碎念:
这是leetcode里面第一个涉及图的题,图应该是数据结构里面的难点,因此这题的成绩很分散。同时说明,即使算法思路相同,内部实现细节不同也会导致成绩有很大差异。比如记录邻接表的时候人家用的是list的下标而不是使用Map的key值,而一直用leaves数组维护当前叶子节点而不是一直在数度,也是比较巧妙的。
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