1002. Find Common Characters

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

这题是找给出的字符串数组的相同的字符。我这里偷懒，直接硬解。找出每个字符串字符出现的次数，然后在别的字符串里找，找到的最小的就是我们要的结果。

class FindCommonCharacters : public Solution {
  public:
    void Execute() {
        auto res = commonChars(vector<string> {"cool"});
    }
    vector<string> commonChars(const vector<string> &A) {
        vector<map<char, int>> counters;

        for (auto &line : A) {

            map<char, int> counter;
            for (auto c : line) {
                auto it = counter.find(c);
                if (it == counter.end()) {
                    counter.insert(std::make_pair(c, 1));
                } else {
                    it->second++;
                }
            }
            counters.push_back(std::move(counter));
        }

        // Loop maps
        vector<string> res;
        if (counters.size() == 1) {
            // Split the word
            for (auto c : A[0]) {
                char ch[2];
                ch[0] = c;
                ch[1] = '\0';
                res.push_back(std::string(ch));
            }
            return res;
        }
        for (int i = 1; i < counters.size(); i++) {
            for (auto &pa : counters[0]) {
                auto it = counters[i].find(pa.first);
                if (it == counters[i].end()) {
                    // Not found in other string
                    pa.second = 0;
                } else {
                    // Select the min one
                    pa.second = pa.second < it->second ? pa.second : it->second;
                }
            }
        }
        for (auto &pa : counters[0]) {
            for (int i = 0; i < pa.second; i++) {
                char ch[2];
                ch[0] = pa.first;
                ch[1] = 0;
                res.push_back(std::string(ch));
            }
        }
        return res;
    }
};