图基础算法（四）- 拓扑排序

There is a new alien language that uses the English alphabet. However, the order among the letters is unknown to you.

You are given a list of strings words from the alien language's dictionary, where the strings in words are sorted lexicographically by the rules of this new language.

Return a string of the unique letters in the new alien language sorted in lexicographically increasing order by the new language's rules. If there is no solution, return "". If there are multiple solutions, return any of them.

A string s is lexicographically smaller than a string t if at the first letter where they differ, the letter in s comes before the letter in t in the alien language. If the first min(s.length, t.length) letters are the same, then s is smaller if and only if s.length < t.length.

Example 1:

Input: words = ["wrt","wrf","er","ett","rftt"]
Output: "wertf"

Example 2:

Input: words = ["z","x"]
Output: "zx"

Example 3:

Input: words = ["z","x","z"]
Output: ""
Explanation: The order is invalid, so return "".

Constraints:

• 1 <= words.length <= 100
• 1 <= words[i].length <= 100
• words[i] consists of only lowercase English letters.

class Solution {
    public String alienOrder(String[] words) {
        //1.build the graph
        Map> graph = new HashMap();
        Map indegree = new HashMap();
        for(String str:words){
            for(char c:str.toCharArray()){
                indegree.put(c,0);
            }
        }
        for(int i=0;i){
            String str1 = words[i],str2=words[i+1];
            for(int j=0;j){
                if(str1.length()>str2.length() && str1.startsWith(str2)) return ""; //坑点1.["abc","ab"] 为不合法的字典，直接返回
                char c1=str1.charAt(j),c2 = str2.charAt(j);
                if(c1!=c2){
                    List list = graph.getOrDefault(c1,new ArrayList());
                    graph.put(c1,list);
                    list.add(c2);
                    indegree.put(c2,indegree.getOrDefault(c2,0)+1);
                    break;
                }
            }
        }
        //2.toplogical sort
        Queue queue = new LinkedList();
        StringBuffer sb = new StringBuffer("");
        for(char c:indegree.keySet()){
            if(indegree.get(c)==0) {
                queue.offer(c);
                sb.append(c);
            }
        }
        while(!queue.isEmpty()){
            char c = queue.poll();
            for(char t:graph.getOrDefault(c,Arrays.asList())){
                indegree.put(t,indegree.get(t)-1);
                if(indegree.get(t)==0) {
                    queue.offer(t);
                    sb.append(t);
                }
            }
        }
        return sb.length()==indegree.size() ? sb.toString() : "";
    }
}