5月挑战Day22-Sort Characters By Frequency（Medium)-优快云博客

本文介绍了一种使用Python解决LeetCode中“按字符频率排序”问题的方法。通过收集字符串中每个字符的频率，然后根据频率降序排序，最后重组字符串。这种方法的时间复杂度为O(nlogn)，空间复杂度为O(n)。

Day22-Sort Characters By Frequency（Medium)

问题描述：

Given a string, sort it in decreasing order based on the frequency of characters.

给定一个字符串，根据字符出现的频率以降序对其进行排序。

Example:

Example 1:

Input:
"tree"

Output:
"eert"

Explanation:
'e' appears twice while 'r' and 't' both appear once.
So 'e' must appear before both 'r' and 't'. Therefore "eetr" is also a valid answer.
Example 2:

Input:
"cccaaa"

Output:
"cccaaa"

Explanation:
Both 'c' and 'a' appear three times, so "aaaccc" is also a valid answer.
Note that "cacaca" is incorrect, as the same characters must be together.
Example 3:

Input:
"Aabb"

Output:
"bbAa"

Explanation:
"bbaA" is also a valid answer, but "Aabb" is incorrect.
Note that 'A' and 'a' are treated as two different characters.

解法：

看到这道题的直观感受就是用一个额外的字典存储每个字符出现的次数，然我们对字典的值进行降序排序，然后再取出排完序的字典的键乘以对应的次数就可以了。

class Solution:
    def frequencySort(self, s: str) -> str:
        #设置一个字典存储每个字符出现的次数
        result_dict = collections.Counter(s)
        print(result_dict)
        sort_list = sorted(result_dict.items(),key = lambda item:item[1],reverse = True)
        print(sort_list)
        result = ''
        for i in sort_list:
            result += i[0] * i[1]
            
        return result