标签：容器 gic leave size result ica 替代 rect efi
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] Note: (1) According to the definition of tree on Wikipedia: “a tree is an undirected graph in which any two vertices are connected by exactly one path. In other words, any connected graph without simple cycles is a tree.” (2) The height of a rooted tree is the number of edges on the longest downward path between the root and a leaf. Credits: Special thanks to @dietpepsi for adding this problem and creating all test cases.
剪枝法, 不能用队, 用剩余节点个数做while循环, 建立叶结点的容器当作队, 需要更新叶结点容器不能用队.
错误的做法: 跟207 Course Schedule有区别, 不能一味套, 因为这是无向图
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
if (n == 1) return Collections.singletonList(0);
List<Integer> leaves = new ArrayList<>();
List[] adj = new ArrayList[n];
int[] indeg = new int[n];
for (int i = 0; i < n; i++) {
adj[i] = new ArrayList<Integer>();
}
for (int[] edge : edges) {
adj[edge[0]].add(edge[1]);
adj[edge[1]].add(edge[0]);
indeg[edge[0]]++;
indeg[edge[1]]++;
}
for (int i = 0; i < n; i++) {
if (indeg[i] == 1) {
leaves.add(i);
}
}
while (n > 2) {
n -= leaves.size();
List<Integer> newList = new ArrayList<>();
for (int i : leaves) {
int j = (int)adj[i].get(0);
indeg[j]--;
if (indeg[j] == 1) {
newList.add(j);
}
}
leaves = newList;
}
return leaves;
}
}
应改为:, 因为get(0) 不一定get到当前叶节点的下一个内靠的叶结点, 可能是已经遍历过得呢
while (n > 2) {
n -= leaves.size();
List<Integer> newList = new ArrayList<>();
for (int i : leaves) {
ArrayList<Integer> list = adj[i];
for (Integer k : list) {
indeg[k]--;
if (indeg[k] == 1) {
newList.add(k);
}
}
}
leaves = newList;
}
跟其最接近的题目是Course Schedule 课程清单和Course Schedule II 课程清单之二。由于LeetCode中的树的题目主要都是针对于二叉树的，而这道题虽说是树但其实本质是想考察图的知识，
用的hashmap 来替代arraylist[]: 这样的好处是方便键不是顺序的时候, 与数组本质是一样的, 但是存储的键更灵活, 应该学会用map
public class Solution {
public List<Integer> findMinHeightTrees(int n, int[][] edges) {
List<Integer> leaves = new ArrayList<Integer>();
if (edges==null || edges.length==0) {
leaves.add(n-1);
return leaves;
}
HashMap<Integer, ArrayList<Integer>> graph = new HashMap<Integer, ArrayList<Integer>>();
int[] indegree = new int[n];
for (int i=0; i<n; i++) {
graph.put(i, new ArrayList<Integer>());
}
//build the graph
for (int[] edge : edges) {
graph.get(edge[0]).add(edge[1]);
graph.get(edge[1]).add(edge[0]);
indegree[edge[0]]++;
indegree[edge[1]]++;
}
//find the leaves
for (int i=0; i<n; i++) {
if (indegree[i] == 1) {
leaves.add(i);
}
}
//topological sort until n<=2
while (n > 2) {
List<Integer> newLeaf = new ArrayList<Integer>();
for (Integer leaf : leaves) {
List<Integer> neighbors = graph.get(leaf);
for (Integer neighbor : neighbors) {
indegree[neighbor]--;
graph.get(neighbor).remove(leaf);
if (indegree[neighbor] == 1)
newLeaf.add(neighbor);
}
//delete leaf from graph
n--;
}
leaves = newLeaf;
}
return leaves;
}
}
标签：容器 gic leave size result ica 替代 rect efi
原文地址：http://www.cnblogs.com/apanda009/p/7226736.html
