LeetCode 916. 单词子集（计数）

示例 1：
输入：A = ["amazon","apple","facebook","google","leetcode"], 
B = ["e","o"]
输出：["facebook","google","leetcode"]

示例 2：
输入：A = ["amazon","apple","facebook","google","leetcode"], 
B = ["l","e"]
输出：["apple","google","leetcode"]

示例 3：
输入：A = ["amazon","apple","facebook","google","leetcode"], 
B = ["e","oo"]
输出：["facebook","google"]

示例 4：
输入：A = ["amazon","apple","facebook","google","leetcode"], 
B = ["lo","eo"]
输出：["google","leetcode"]

示例 5：
输入：A = ["amazon","apple","facebook","google","leetcode"], 
B = ["ec","oc","ceo"]
输出：["facebook","leetcode"]
 
提示：
1 <= A.length, B.length <= 10000
1 <= A[i].length, B[i].length <= 10
A[i] 和 B[i] 只由小写字母组成。
A[i] 中所有的单词都是独一无二的，也就是说不存在 i != j 使得 A[i] == A[j]。

class Solution {
public:
    vector<string> wordSubsets(vector<string>& A, vector<string>& B) {
    	vector<int> countB(26,0);
    	for(auto& bi : B)
    	{
            vector<int> countb(26,0);
            for(char ch : bi)
    			countb[ch-'a']++;
            for(int i = 0; i < 26; ++i)
                countB[i] = max(countB[i],countb[i]);//取较大的
        }

    	vector<string> ans;
    	bool ok;
    	for(auto& Ai : A)
    	{
            vector<int> countai(26,0);
    		ok = true;
    		for(char ch : Ai)
                countai[ch-'a']++;
    		for(int i = 0; i < 26; ++i)
            {
                if(countai[i] < countB[i])//不满足
                {
                    ok = false;
                    break;
                }
            }
            if(ok)
    			ans.push_back(Ai);
    	}
    	return ans;
    }
};