[LeetCode] 1002. Find Common Characters 尋找相同字符

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

這道題給了一組由小寫字母組成的字符串，讓返回所有的字符串中的相同的字符，重復的字符要出現對應的次數。其實這道題的核心就是如何判斷字符串的相交部分，跟之前的 Intersection of Two Arrays II 和 Intersection of Two Arrays 比較類似。核心是要用 HashMap 建立字符和其出現次數之間的映射，這里由於只有小寫字母，所以可以使用一個大小為 26 的數組來代替 HashMap。用一個數組 cnt 來記錄相同的字母出現的次數，初始化為整型最大值，然后遍歷所有的單詞，對於每個單詞，新建一個大小為 26 的數組t，並統計每個字符出現的次數，然后遍歷0到25各個位置，取 cnt 和 t 對應位置上的較小值來更新 cnt 數組，這樣得到就是在所有單詞里都出現的字母的個數，最后再把這些字符轉為字符串加入到結果 res 中即可，參見代碼如下：

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:

https://github.com/grandyang/leetcode/issues/1002

類似題目：

Intersection of Two Arrays II

Intersection of Two Arrays

參考資料：

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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