We are given two arrays A and B of words.  Each word is a string of lowercase letters.

Now, say that word b is a subset of word a if every letter in b occurs in a, including multiplicity.  For example, "wrr" is a subset of "warrior", but is not a subset of "world".

Now say a word a from A is  universal  if for every b in B, b is a subset of a. 

Return a list of all universal words in A.  You can return the words in any order.

Example 1:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","o"]
Output: ["facebook","google","leetcode"]

Example 2:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["l","e"]
Output: ["apple","google","leetcode"]

Example 3:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["e","oo"]
Output: ["facebook","google"]

Example 4:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["lo","eo"]
Output: ["google","leetcode"]

Example 5:

Input: A = ["amazon","apple","facebook","google","leetcode"], B = ["ec","oc","ceo"]
Output: ["facebook","leetcode"]

Note:

1. 1 <= A.length, B.length <= 10000
2. 1 <= A[i].length, B[i].length <= 10
3. A[i] and B[i] consist only of lowercase letters.
4. All words in A[i] are unique: there isn't i != j with A[i] == A[j].

这道题定义了两个单词之间的一种子集合关系，就是说假如单词b中的每个字母都在单词a中出现了（包括重复字母），就说单词b是单词a的子集合。现在给了两个单词集合A和B，让找出集合A中的所有满足要求的单词，使得集合B中的所有单词都是其子集合。配合上题目中给的一堆例子，意思并不难理解，根据子集合的定义关系，其实就是说若单词a中的每个字母的出现次数都大于等于单词b中每个字母的出现次数，单词b就一定是a的子集合。现在由于集合B中的所有单词都必须是A中某个单词的子集合，那么其实只要对于每个字母，都统计出集合B中某个单词中出现的最大次数，比如对于这个例子，B=["eo","oo"]，其中e最多出现1次，而o最多出现2次，那么只要集合A中有单词的e出现不少1次，o出现不少于2次，则集合B中的所有单词一定都是其子集合。这就是本题的解题思路，这里使用一个大小为 26 的一维数组 charCnt 来统计集合B中每个字母的最大出现次数，而将统计每个单词的字母次数的操作放到一个子函数 helper 中，当 charCnt 数组更新完毕后，下面就开始检验集合A中的所有单词了。对于每个遍历到的单词，还是要先统计其每个字母的出现次数，然后跟 charCnt 中每个位置上的数字比较，只要均大于等于 charCnt 中的数字，就可以加入到结果 res 中了，参见代码如下：

class Solution {
public:
    vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
        vector<string> res;
        vector<int> charCnt(26);
        for (string &b : B) {
            vector<int> t = helper(b);
            for (int i = 0; i < 26; ++i) {
                charCnt[i] = max(charCnt[i], t[i]);
            }
        }
        for (string &a : A) {
            vector<int> t = helper(a);
            int i = 0;
            for (; i < 26; ++i) {
                if (t[i] < charCnt[i]) break;
            }
            if (i == 26) res.push_back(a);
        }
        return res;
    }
    vector<int> helper(string& word) {
        vector<int> res(26);
        for (char c : word) ++res[c - 'a'];
        return res;
    }
};

Github 同步地址:

#916

参考资料：

https://leetcode.com/problems/word-subsets/

https://leetcode.com/problems/word-subsets/discuss/175854/C%2B%2BJavaPython-Straight-Forward

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