# 1002.Find Common Characters

**1002.Find Common Characters**

难度:Easy

> 给定仅有小写字母组成的字符串数组 A，返回列表中的每个字符串中都显示的全部字符（包括重复字符）组成的列表。例如，如果一个字符在每个字符串中出现 3 次，但不是 4 次，则需要在最终答案中包含该字符 3 次。

你可以按任意顺序返回答案。

```
示例 1：

输入：["bella","label","roller"]
输出：["e","l","l"]
示例 2：

输入：["cool","lock","cook"]
输出：["c","o"]
 

提示：

1 <= A.length <= 100
1 <= A[i].length <= 100
A[i][j] 是小写字母
```

方法：将每个单词字母用字典存起来，比较出现次数，取出现次数少的保存。

```
class Solution {
public:
    vector<string> commonChars(vector<string>& A) {
        vector<string> res;
        unordered_map<char,int> cnt;

        int len=A.size();
        for(auto c: A[0])
            cnt[c]++;
        for(int i=1;i<len;i++)
        {
            unordered_map<char,int> tmp;
            for(auto c:A[i])
                tmp[c]++ ;
            for(char c='a';c<='z';c++)
               // cout<<cnt[c]<<c<<tmp[c]<<endl;
                cnt[c]=min(tmp[c],cnt[c]);
        }
        for(char c='a';c<='z';c++)
            while(cnt[c]--)
                res.push_back(string(1,c));
        return res;
    }
};
```

> 执行用时 : 84 ms, 在Find Common Characters的C++提交中击败了100.00% 的用户 内存消耗 : 14.1 MB, 在Find Common Characters的C++提交中击败了100.00% 的用户


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