<p>Given a string array <code>words</code>, return <em>an array of all characters that show up in all strings within the </em><code>words</code><em> (including duplicates)</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> words = ["bella","label","roller"]
<strong>Output:</strong> ["e","l","l"]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> words = ["cool","lock","cook"]
<strong>Output:</strong> ["c","o"]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 100</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    vector<string> commonChars(vector<string>& words) {
        vector<string> result;
        if (words.size() == 0) return result;
        int hash[26] = {0}; // 用来统计所有字符串里字符出现的最小频率
        for (int i = 0; i < words[0].size(); i++) { // 用第一个字符串给hash初始化
            hash[words[0][i] - 'a']++;
        }

        int hashOtherStr[26] = {0}; // 统计除第一个字符串外字符的出现频率
        for (int i = 1; i < words.size(); i++) {
            memset(hashOtherStr, 0, 26 * sizeof(int));
            for (int j = 0; j < words[i].size(); j++) {
                hashOtherStr[words[i][j] - 'a']++;
            }
            // 更新hash，保证hash里统计26个字符在所有字符串里出现的最小次数
            for (int k = 0; k < 26; k++) {
                hash[k] = min(hash[k], hashOtherStr[k]);
            }
        }
        // 将hash统计的字符次数，转成输出形式
        for (int i = 0; i < 26; i++) {
            while (hash[i] != 0) { // 注意这里是while，多个重复的字符
                string s(1, i + 'a'); // char -> string
                result.push_back(s);
                hash[i]--;
            }
        }

        return result;
    }
};
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