从给定原材料中找到所有可以做出的菜

/*
 * 题目思路：
 * 1. 使用Java Stream API实现拓扑排序的逻辑。
 * 2. 使用集合保存初始的supplies。
 * 3. 通过stream和filter筛选可以制作的菜。
 */

import java.util.*;
import java.util.stream.Collectors;

public class RecipeFinderStream {
    public List<String> findAllRecipes(String[] recipes, List<List<String>> ingredients, String[] supplies) {
        Set<String> supplySet = new HashSet<>(Arrays.asList(supplies));
        Map<String, Long> indegree = new HashMap<>();
        Map<String, List<String>> graph = new HashMap<>();

        Arrays.stream(recipes).forEach(recipe -> indegree.put(recipe, 0L));

        for (int i = 0; i < recipes.length; i++) {
            final int index = i;
            ingredients.get(i).stream()
                .filter(ingredient -> !supplySet.contains(ingredient))
                .forEach(ingredient -> {
                    graph.computeIfAbsent(ingredient, x -> new ArrayList<>()).add(recipes[index]);
                    indegree.put(recipes[index], indegree.get(recipes[index]) + 1);
                });
        }

        return indegree.keySet().stream()
            .filter(recipe -> indegree.get(recipe) == 0)
            .flatMap(recipe -> {
                Queue<String> queue = new LinkedList<>();
                List<String> result = new ArrayList<>();
                queue.offer(recipe);
                while (!queue.isEmpty()) {
                    String current = queue.poll();
                    result.add(current);
                    if (graph.containsKey(current)) {
                        graph.get(current).stream()
                            .filter(next -> indegree.put(next, indegree.get(next) - 1) == 1)
                            .forEach(queue::offer);
                    }
                }
                return result.stream();
            })
            .collect(Collectors.toList());
    }
}