Given an array A of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates).  For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.

You may return the answer in any order.

Example 1:

Input: ["bella","label","roller"]Output: ["e","l","l"]

Example 2:

Input: ["cool","lock","cook"]Output: ["c","o"]

Note:

1. 1 <= A.length <= 100
2. 1 <= A[i].length <= 100
3. A[i][j] is a lowercase letter

这道题给了一组由小写字母组成的字符串，让返回所有的字符串中的相同的字符，重复的字符要出现对应的次数。其实这道题的核心就是如何判断字符串的相交部分，跟之前的Intersection of Two Arrays II 和Intersection of Two Arrays 比较类似。核心是要用 HashMap 建立字符和其出现次数之间的映射，这里由于只有小写字母，所以可以使用一个大小为 26 的数组来代替 HashMap。用一个数组 cnt 来记录相同的字母出现的次数，初始化为整型最大值，然后遍历所有的单词，对于每个单词，新建一个大小为 26 的数组t，并统计每个字符出现的次数，然后遍历0到25各个位置，取 cnt 和 t 对应位置上的较小值来更新 cnt 数组，这样得到就是在所有单词里都出现的字母的个数，最后再把这些字符转为字符串加入到结果 res 中即可，参见代码如下：

class Solution {public:    vector<string> commonChars(vector<string>& A) {        vector<string> res;        vector<int> cnt(26, INT_MAX);        for (string word : A) {            vector<int> t(26);            for (char c : word) ++t[c - 'a'];            for (int i = 0; i < 26; ++i) {                cnt[i] = min(cnt[i], t[i]);            }        }        for (int i = 0; i < 26; ++i) {            for (int j = 0; j < cnt[i]; ++j) {                res.push_back(string(1, 'a' + i));            }        }        return res;    }};

Github 同步地址:

#1002

类似题目：

Intersection of Two Arrays II

Intersection of Two Arrays

参考资料：

https://leetcode.com/problems/find-common-characters/

https://leetcode.com/problems/find-common-characters/discuss/247573/C%2B%2B-O(n)-or-O(1)-two-vectors

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