查找表_leetcode451

# 解题思路：字典存储计数信息 20190302 找工作期间

class Solution(object):
    def frequencySort(self, s):
        """
        :type s: str
        :rtype: str
        """

        dict1 = {}
        result = str()
        for i in range(len(s)):
            if s[i] not in dict1:
                dict1[s[i]] = 1
            else:
                dict1[s[i]] += 1
        while len(dict1) > 0:
            m = max(dict1, key=dict1.get)
            result += m * dict1[m]
            del dict1[m]
        return result

查找表_leetcode451

原文：https://www.cnblogs.com/lux-ace/p/10546967.html