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].

我们给出两个单词数组 A 和 B。每个单词都是一串小写字母。现在，如果 b 中的每个字母都出现在 a 中，包括重复出现的字母，那么称单词 b 是单词 a 的子集。 例如，“wrr” 是 “warrior” 的子集，但不是 “world” 的子集。如果对 B 中的每一个单词 b，b 都是 a 的子集，那么我们称 A 中的单词 a 是通用的。你可以按任意顺序以列表形式返回 A 中所有的通用单词。

• 简单题。先统计出 B 数组中单词每个字母的频次，再在 A 数组中依次判断每个单词是否超过了这个频次，如果超过了即输出。

package leetcode funcwordSubsets(A []string, B []string) []string { varcounter [26]intfor_, b:=rangeB { varm [26]intfor_, c:=rangeb { j:=c-'a'm[j]++ } fori:=0; i<26; i++ { ifm[i] >counter[i] { counter[i] =m[i] } } } varres []stringfor_, a:=rangeA { varm [26]intfor_, c:=rangea { j:=c-'a'm[j]++ } ok:=truefori:=0; i<26; i++ { ifm[i] <counter[i] { ok=falsebreak } } ifok { res=append(res, a) } } returnres }