[leetcode] 题型整理之图论

图论的常见题目有两类，一类是求两点间最短距离，另一类是拓扑排序，两种写起来都很烦。

求最短路径：

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:

1. Only one letter can be changed at a time
2. 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.

从起点开始向外更新，因为每条路径的权值都不是负数，所以先更新的总比后更新的小。

已经被更新过的之后就不用考虑了

 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 #.

1. First node is labeled as 0. Connect node 0 to both nodes 1 and 2.
2. Second node is labeled as 1. Connect node 1 to node 2.
3. 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
         / 
         \_/

关键要用一个图把旧的节点和新的节点一一对应起来，并用一个队列存储需要更新neighbor的节点，每次加入到某一个链表中的新节点，只在map中放入neighbor没有更新的新节点，待从queue中读到的时候再处理。

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:

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

1. 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;
    }
}