451. 根据字符出现频率排序

class Solution(object):
    def frequencySort(self, s):
        """
        :type s: str
        :rtype: str
        """
        mydict = {}
        for item in s:
            if item in mydict.keys():
                mydict[item] += 1
            else:
                mydict[item] = 1
        newlist = sorted(mydict.items(), key=lambda x: x[1], reverse=True)
        res = ""
        for i in newlist:
            res += "".join(i[0] * i[1])
        return res

451. 根据字符出现频率排序

原文：https://www.cnblogs.com/panweiwei/p/14024883.html