題目描述：
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 <= A.length, B.length <= 100001 <= A[i].length, B[i].length <= 10A[i]andB[i]consist only of lowercase letters.All words in
A[i]are unique: there isn’ti != jwithA[i] == A[j].
一刷題解（Brute Force）：
這題給了我們兩個字符串數組，A字符串數組裡的都是一些完整的字符串，而B字符串數組裡則都是一些字符碎片，題目要我們在A字符串數組中找到一些字符串，這些字符串中各類字母，包含且大於B字符串中的各類字母出現的最大數量。
因此，我們可以先統計B字符串數組中，各類字母出現的最大次數，並將其放到一個容量為26的數組裡（發揮字母出現次數字典的功能）。完成B的統計後，我們再逐一統計每個A字符串裡，出現的字母是否包括了B字符串的字母，出現次數是否大於B字符串字母的出現次數，如果兩者答案均為肯定，那就代表這個A字符串是我們的其中一個答案。
public class Solution {
public IList<string> WordSubsets(string[] A, string[] B)
{
List<string> result = new List<string>();
//字母出現統計數組
int[] charAppearCnt = new int[26];
//統計B中的字符裡的每類字母出現的最大數量
for (int i = 0; i < B.Length; i++)
{
int[] cntArr = CharCount(B[i]);
for (int j = 0; j < cntArr.Length; j++)
{
if (cntArr[j] == 0) { continue; }
if (charAppearCnt[j] < cntArr[j])
{
charAppearCnt[j] = cntArr[j];
}
}
}
//檢查A中的字符的各類字母是包含且大於字典裡的記錄
for (int i = 0; i < A.Length; i++)
{
bool isUniversal = true;
int[] cntArr = CharCount(A[i]);
for (int j = 0; j < cntArr.Length; j++)
{
//只需要檢查B字符串裡存在的字母，其他沒出現過的字母跳過檢查
if (charAppearCnt[j] == 0) { continue; }
//如果A中字母出現次數少於B字符字母的出現次數，該字符B就不是字符A的Subset
if (cntArr[j] < charAppearCnt[j])
{
isUniversal = false;
break;
}
}
if (isUniversal)
{
result.Add(A[i]);
}
}
return result;
}
int[] CharCount(string str)
{
int[] cntArr = new int[26];
foreach (char c in str.ToCharArray())
{
cntArr[c - 'a'] += 1;
}
return cntArr;
}
}