1452. People Whose List of Favorite Companies Is Not a Subset of Another List

Given the array favoriteCompanies where favoriteCompanies[i] is the list of favorites companies for the ith person (indexed from 0).

Return the indices of people whose list of favorite companies is not a subset of any other list of favorites companies. You must return the indices in increasing order.

Example 1:

Input: favoriteCompanies = [["leetcode","google","facebook"],["google","microsoft"],["google","facebook"],["google"],["amazon"]]
Output: [0,1,4] 
Explanation: 
Person with index=2 has favoriteCompanies[2]=["google","facebook"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] corresponding to the person with index 0. 
Person with index=3 has favoriteCompanies[3]=["google"] which is a subset of favoriteCompanies[0]=["leetcode","google","facebook"] and favoriteCompanies[1]=["google","microsoft"]. 
Other lists of favorite companies are not a subset of another list, therefore, the answer is [0,1,4].

Example 2:

Input: favoriteCompanies = [["leetcode","google","facebook"],["leetcode","amazon"],["facebook","google"]]
Output: [0,1] 
Explanation: In this case favoriteCompanies[2]=["facebook","google"] is a subset of favoriteCompanies[0]=["leetcode","google","facebook"], therefore, the answer is [0,1].

Example 3:

Input: favoriteCompanies = [["leetcode"],["google"],["facebook"],["amazon"]]
Output: [0,1,2,3]

Constraints:

• 1 <= favoriteCompanies.length <= 100
• 1 <= favoriteCompanies[i].length <= 500
• 1 <= favoriteCompanies[i][j].length <= 20
• All strings in favoriteCompanies[i] are distinct.
• All lists of favorite companies are distinct, that is, If we sort alphabetically each list then favoriteCompanies[i] != favoriteCompanies[j].
• All strings consist of lowercase English letters only.

class Solution {
    public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
        Set<String>[] sets = new HashSet[favoriteCompanies.size()];
        for(int i = 0; i < sets.length; i++) sets[i] = new HashSet(favoriteCompanies.get(i));
        List<Integer> res = new ArrayList();
        
        for(int i = 0; i < sets.length; i++){
            boolean flag = true;
            for(int j = 0; j < sets.length; j++){
                if(j != i && sets[i].size() < sets[j].size() && sets[j].containsAll(sets[i])){
                    flag = false; 
                    break;
                }
            }
            if(flag == true) res.add(i);
        }
        return res;
    }
}

利用了set的containsAll method，检查setA是否包含了所有setB的元素。

或者

class Solution {
     public List<Integer> peopleIndexes(List<List<String>> favoriteCompanies) {
            Set<String>[] set = new Set[favoriteCompanies.size()];
            for (int i = 0; i < set.length; ++i)
                set[i] = new HashSet<>(favoriteCompanies.get(i));
            List<Integer> res = new ArrayList<>();
            outer: 
            for (int i = 0; i < set.length; ++i) {
                for (int j = 0; j < set.length; ++j)
                    if (i != j && set[j].containsAll(set[i]))
                        continue outer;
                res.add(i);
            }
            return res;        
        }
}

outer关键词可以帮助跳出外循环，具体用法如下

https://blog.csdn.net/qq_41668547/article/details/80527969?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-2.nonecase&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-2.nonecase