LeetCode | 0451. Sort Characters By Frequency根据字符出现频率排序【Python】
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2022-05-13 21:25:48
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LeetCode 0451. Sort Characters By Frequency根据字符出现频率排序【Medium】【Python】【桶排序】
Problem
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.
问题
给定一个字符串,请将字符串里的字符按照出现的频率降序排列。
示例 1:
输入:
"tree"
输出:
"eert"
解释:
'e'出现两次,'r'和't'都只出现一次。
因此'e'必须出现在'r'和't'之前。此外,"eetr"也是一个有效的答案。
示例 2:
输入:
"cccaaa"
输出:
"cccaaa"
解释:
'c'和'a'都出现三次。此外,"aaaccc"也是有效的答案。
注意"cacaca"是不正确的,因为相同的字母必须放在一起。
示例 3:
输入:
"Aabb"
输出:
"bbAa"
解释:
此外,"bbaA"也是一个有效的答案,但"Aabb"是不正确的。
注意'A'和'a'被认为是两种不同的字符。
思路
桶排序
统计字符串字符出现的次数,然后桶排序,最后逆序输出字符串。
可以直接用 Python 的Counter类就能统计每个字符出现的次数,使用most_common函数就能按次序排列。
时间复杂度: O(n)
空间复杂度: O(m) m是桶的数量
Python代码
class Solution:
def frequencySort(self, s):
"""
:type s: str
:rtype: str
"""
# # 法一:collections.Counter
# import collections # 需要导入collections
# count = collections.Counter(s).most_common() # Counter类就能统计每个字符出现的次数, 使用most_common函数就能按次序排列
# res = ''
# for char, cnt in count:
# res += char*cnt # 字符与次数相乘
# return res
# 法二:桶排序
import collections
# 统计字符个数
ret = []
count_frequency = collections.defaultdict(int)
for i in s:
count_frequency[i] += 1 # 统计字符个数
# 桶排序
buckets = [[] for i in range(len(s) + 1)]
for i in count_frequency: # i是字符
# print (i)
buckets[count_frequency[i]].extend(i * count_frequency[i])
# 打印排完序的字符串
for i in buckets[::-1]: # -1表示逆序
if(i):
ret.extend(i)
return ''.join(ret)
# return ''.join(c*-n for n,c in sorted((-s.count(c),c)for c in set(s))) # 一行代码也能解决