leetcode

'''
个人理解的作者的思路：将字符重合问题使用字典来记录每个char出现的次数，遍历整个list，每次遍历都会求出中间结果（commonCharArray），中间结果的求法是对比当前字典（wordCharDict）与上一次字典（orderedDict）的记录。
'''
class :
    def commonChars(self, A):
        orderedDict = {}
        for _, word in enumerate(A):
            wordCharDict = {} 
            for char in word:
                if char in wordCharDict:
                    value = wordCharDict[char]
                    wordCharDict[char] = value + 1
                else:
                    wordCharDict[char] = 1

            commonCharArray = [] # 保存公共的char
            # 对orderedDict初始化
            if not bool(orderedDict):
                orderedDict = wordCharDict

            # orderedDict保存了当前公共的char及其相应的出现次数
            for char, occurence in orderedDict.items():
                commonOcurrence = 0
                if char in wordCharDict:
                    commonOcurrence = min(occurence, wordCharDict[char])
                orderedDict[char] = commonOcurrence
                
                # 将当前为止公共的char保存起来，当最后一次遍历结束，即为最终结果
                for _ in range(commonOcurrence):
                    commonCharArray.append(char)
        return commonCharArray