[leetcode] 451. Sort Characters By Frequency @ python

最新推荐文章于 2025-02-03 15:37:10 发布

原创最新推荐文章于 2025-02-03 15:37:10 发布 · 386 阅读

0 ·

CC 4.0 BY-SA版权

本文介绍了一种基于字符出现频率的字符串排序算法。通过将字符串转化为集合并构建字典，字典的键为频率，值为对应字符列表，实现对字符串中字符按频率从高到低排序。代码使用Python实现，时间复杂度为O(n)，空间复杂度也为O(n)。

原题

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

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

将字符串转化为集合以减少遍历的次数, 构造字典, 字典的键是频率, 值是对应的字母列表, 然后降序排列频率, 遍历频率, 将频率对应的字母加到ans里.
Time: O(n), n为s中不同字母的个数
Space: O(n)

代码

class Solution:
    def frequencySort(self, s: 'str') -> 'str':
        str_s = set(s)
        d = collections.defaultdict(list)
        for char in str_s:
            count = s.count(char)
            d[count].append(char)
            
        ans = ''
        
        for k in sorted(d.keys(), reverse = True):
            while len(d[k]) > 0:
                char = d[k].pop()
                ans += char*k
        return ans