圖論的常見題目有兩類，一類是求兩點(diǎn)間最短距離，另一類是拓?fù)渑判?#xff0c;兩種寫起來都很煩。

求最短路徑：

127. Word Ladder

Given two words (beginWord?and?endWord), and a dictionary's word list, find the length of shortest transformation sequence from?beginWord?to?endWord, such that:

Only one letter can be changed at a time

Each intermediate word must exist in the word list

For example,

Given:
beginWord?=?"hit"
endWord?=?"cog"
wordList?=?["hot","dot","dog","lot","log"]

As one shortest transformation is?"hit" -> "hot" -> "dot" -> "dog" -> "cog",
return its length?5.

Note:

• • Return 0 if there is no such transformation sequence.
  • All words have the same length.

All words contain only lowercase alphabetic characters.

從起點(diǎn)開始向外更新，因?yàn)槊織l路徑的權(quán)值都不是負(fù)數(shù)，所以先更新的總比后更新的小。

已經(jīng)被更新過的之后就不用考慮了

?133. Clone Graph

Clone an undirected graph. Each node in the graph contains a?label?and a list of its?neighbors.

OJ's undirected graph serialization:

Nodes are labeled uniquely.

We use?#?as a separator for each node, and?,?as a separator for node label and each neighbor of the node.

?

As an example, consider the serialized graph?{0,1,2#1,2#2,2}.

The graph has a total of three nodes, and therefore contains three parts as separated by?#.

First node is labeled as?0. Connect node?0?to both nodes?1?and?2.

Second node is labeled as?1. Connect node?1?to node?2.

Third node is labeled as?2. Connect node?2?to node?2?(itself), thus forming a self-cycle.

?

Visually, the graph looks like the following:

1/ \/ \0 --- 2/ \\_/

?

?

關(guān)鍵要用一個圖把舊的節(jié)點(diǎn)和新的節(jié)點(diǎn)一一對應(yīng)起來，并用一個隊(duì)列存儲需要更新neighbor的節(jié)點(diǎn)，每次加入到某一個鏈表中的新節(jié)點(diǎn)，只在map中放入neighbor沒有更新的新節(jié)點(diǎn)，待從queue中讀到的時候再處理。 261. Graph Valid Tree

Given?n?nodes labeled from?0?to?n - 1?and a list of undirected edges (each edge is a pair of nodes), write a function to check whether these edges make up a valid tree.

For example:

Given?n = 5?and?edges = [[0, 1], [0, 2], [0, 3], [1, 4]], return?true.

Given?n = 5?and?edges = [[0, 1], [1, 2], [2, 3], [1, 3], [1, 4]], return?false.

Hint:

Given?n = 5?and?edges = [[0, 1], [1, 2], [3, 4]], what should your return? Is this case a valid tree?

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.”

Note: 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.

public class Solution {public boolean validTree(int n, int[][] edges) {List<HashSet<Integer>> sets = new ArrayList<HashSet<Integer>>();for (int i = 0; i < n; i++) {sets.add(new HashSet<Integer>());}boolean[] isVisited = new boolean[n];Arrays.fill(isVisited, false);for (int[] edge : edges) {int v1 = edge[0];int v2 = edge[1];sets.get(v1).add(v2);sets.get(v2).add(v1);}boolean result = dfs(sets, isVisited, 0, -1);if (!result) {return false;}for (int i = 0; i < n; i++) {if (!isVisited[i]) {return false;}}return true;}private boolean dfs(List<HashSet<Integer>> sets, boolean[] isVisited, int now, int prev) {boolean isV = isVisited[now];if (isV) {return false;}isVisited[now] = true;for (Integer x : sets.get(now)) {if (x != prev && !dfs(sets, isVisited, x, now)) {return false;}}return true;} }

?

310. Minimum Height Trees

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]

Hint:

How many MHTs can a graph have at most?

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.

public class Solution {public List<Integer> findMinHeightTrees(int n, int[][] edges) {List<List<Integer>> graph = new ArrayList<List<Integer>>(n);if (n < 3 || edges.length == 0) {List<Integer> result = new ArrayList<Integer>();if (n == 0) {return result;} else {for (int i = 0; i < n; i++) {result.add(i);} return result;}}for (int i = 0; i < n; i++) {List<Integer> list = new LinkedList<Integer>();graph.add(list);}for (int[] edge : edges) {int v1 = edge[0];int v2 = edge[1];graph.get(v1).add(v2);graph.get(v2).add(v1);}int count = n;List<Integer> toRemove = new ArrayList<Integer>();for (int i = 0; i < n; i++) {List<Integer> list = graph.get(i);if (list.size() == 1) {toRemove.add(i);}}while (!toRemove.isEmpty() && count > 1) {List<Integer> tmpRemove = new ArrayList<Integer>();for (Integer leave : toRemove) {List<Integer> list0 = graph.get(leave);int parent = list0.get(0);list0.clear();List<Integer> list = graph.get(parent);list.remove(leave);if (list.size() == 1) {tmpRemove.add(parent);}}count -= toRemove.size();toRemove = tmpRemove;if (count <= 2) {break;}}List<Integer> result = new ArrayList<Integer>();result.addAll(toRemove);return result;} }

?

轉(zhuǎn)載于:https://www.cnblogs.com/Gryffin/p/6232520.html

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