【leetcode】1002. Find Common Characters

题目如下：

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

解题思路：创建一个char_count[26]数字，记录a-z每一个字符在所有字符串中出现的最少的次数，初始值为101。接下来遍历Input中的每个单词，并且把每个字符在这个单词出现的次数与char_count中对应字符的次数进行比较取较小值。Input遍历完成后，char_count中每个字符所对应的值即为公共的个数。

代码如下：

class Solution(object):
    def commonChars(self, A):
        """
        :type A: List[str]
        :rtype: List[str]
        """
        cl = [101] * 26

        for word in A:
            for char in range(97,97+26):
                inx = char - ord('a')
                cl[inx] = min(cl[inx],word.count(chr(char)))
        res = []
        for i in range(len(cl)):
            res += [chr(i + ord('a'))] * cl[i]
        return res