如何将库存食材列表与嵌套结构的菜谱食材列表比对？

DevOps 小满

2026-8-14

解决方案：库存食材与菜谱匹配

核心逻辑拆解

遍历所有菜谱，检查该菜谱的全部食材是否都存在于库存列表中
统一字符串大小写，避免因大小写差异导致匹配失败
先修正代码里的拼写错误（比如appel应为apple、potatato应为potato，否则会影响匹配结果）

具体实现步骤

将库存食材转成小写集合，提升后续查找效率
遍历菜谱列表，对每个菜谱的食材列表做检查
判断菜谱食材是否全部包含在库存中，筛选出可制作的菜谱

完整代码示例

List<RecipeList> recipesList = [
  RecipeList(recipeName: "Cake", itemNames: ["banana", "apple"]), // 修正拼写：appel → apple
  RecipeList(recipeName: "Soup", itemNames: ["potato", "egg"]), // 修正拼写：potatato → potato
  RecipeList(recipeName: "Sandwich", itemNames: ["Toast", "Sausage", "Ketchup"]),
  RecipeList(recipeName: "Pizza", itemNames: ["Tomato", "Mushroom"]),
];

List<String> inventory = ["coke", "egg", "banana", "apple"];

class RecipeList {
  String? recipeName;
  List<String>? itemNames;

  RecipeList({this.recipeName, this.itemNames});
}

void main() {
  // 转小写集合，快速查找
  final lowerCaseInventory = inventory.map((item) => item.toLowerCase()).toSet();

  // 筛选可制作的菜谱
  List<RecipeList> cookableRecipes = recipesList.where((recipe) {
    if (recipe.itemNames == null) return false;
    // 菜谱食材转小写后，检查是否全部在库存中
    return recipe.itemNames!.every((item) => lowerCaseInventory.contains(item.toLowerCase()));
  }).toList();

  // 输出结果
  print("可制作的菜谱：");
  for (var recipe in cookableRecipes) {
    print("- ${recipe.recipeName}");
  }
}