[leetcode/lintcode 题解]算法面试真题详解：外星人字典

描述
有一种新的使用拉丁字母的外来语言。但是，你不知道字母之间的顺序。你会从词典中收到一个非空的单词列表，其中的单词在这种新语言的规则下按字典顺序排序。请推导出这种语言的字母顺序。

1. 你可以假设所有的字母都是小写。
2. 如果a是b的前缀且b出现在a之前，那么这个顺序是无效的。
3. 如果顺序是无效的，则返回空字符串。
4. 这里可能有多个有效的字母顺序，返回以正常字典顺序看来最小的。

在线评测地址：领扣题库官网

样例1
输入：["wrt","wrf","er","ett","rftt"]
输出："wertf"
解释：
从 "wrt"和"wrf" ,我们可以得到 't'<'f'
从 "wrt"和"er" ,我们可以得到'w'<'e'
从 "er"和"ett" ,我们可以得到 get 'r'<'t'
从 "ett"和"rftt" ,我们可以得到 'e'<'r'
所以返回 "wertf"

样例2
输入：["z","x"]
输出："zx"
解释：
从 "z" 和 "x"，我们可以得到 'z' < 'x'
所以返回"zx"

题解
使用拓扑排序算法。
这个版本的实现当中，无需假设所有字母为小写字母

/**
* This reference program is provided by @jiuzhang.com
* Copyright is reserved. Please indicate the source for forwarding
*/
class Solution {
    public String alienOrder(String[] words) {
        Map<Character, Set<Character>> graph = constructGraph(words);
        return topologicalSorting(graph);
    }
    
        
    private Map<Character, Set<Character>> constructGraph(String[] words) {
        Map<Character, Set<Character>> graph = new HashMap<>();
        
        // create nodes
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words[i].length(); j++) {
                char c = words[i].charAt(j);
                if (!graph.containsKey(c)) {
                    graph.put(c, new HashSet<Character>());
                }
            }
        }
        
        // create edges
        for (int i = 0; i <  words.length - 1; i++) {
            int index = 0;
            while (index < words[i].length() && index < words[i + 1].length()) {
                if (words[i].charAt(index) != words[i + 1].charAt(index)) {
                    graph.get(words[i].charAt(index)).add(words[i + 1].charAt(index));
                    break;
                }
                index++;
            }
        }

        return graph;
    }
    
    private Map<Character, Integer> getIndegree(Map<Character, Set<Character>> graph) {
        Map<Character, Integer> indegree = new HashMap<>();
        for (Character u : graph.keySet()) {
            indegree.put(u, 0);
        }
        
        for (Character u : graph.keySet()) {
            for (Character v : graph.get(u)) {
                indegree.put(v, indegree.get(v) + 1);
            }
        }     
        
        return indegree;
    }

    private String topologicalSorting(Map<Character, Set<Character>> graph) {
        Map<Character, Integer> indegree = getIndegree(graph);
        // as we should return the topo order with lexicographical order
        // we should use PriorityQueue instead of a FIFO Queue
        Queue<Character> queue = new PriorityQueue<>();
        
        for (Character u : indegree.keySet()) {
            if (indegree.get(u) == 0) {
                queue.offer(u);
            }
        }
        
        StringBuilder sb = new StringBuilder();
        while (!queue.isEmpty()) {
            Character head = queue.poll();
            sb.append(head);
            for (Character neighbor : graph.get(head)) {
                indegree.put(neighbor, indegree.get(neighbor) - 1);
                if (indegree.get(neighbor) == 0) {
                    queue.offer(neighbor);
                }
            }
        }
        
        if (sb.length() != indegree.size()) {
            return "";
        }
        return sb.toString();
    }
}

更多题解参考：九章官网solution
https://www.jiuzhang.com/solution/alien-dictionary/?utm_source=sc-zhihuzl-sy