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451. Sort Characters By Frequency ​

题目 ​

Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string.

Return the sorted string. If there are multiple answers, return any of them.

 

Example 1:

Input: s = "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: s = "cccaaa"
Output: "aaaccc"
Explanation: Both 'c' and 'a' appear three times, so both "cccaaa" and "aaaccc" are valid answers.
Note that "cacaca" is incorrect, as the same characters must be together.

Example 3:

Input: s = "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.

 

Constraints:

  • 1 <= s.length <= 5 * 105
  • s consists of uppercase and lowercase English letters and digits.
Related Topics
  • 哈希表
  • 字符串
  • 桶排序
  • 计数
  • 排序
  • 堆（优先队列）

  • 👍 550
  • 👎 0

    • 思路 ​

    将字符串通过dict存储，再通过 python 中的 lambda 表达式进行值排序，和最后的结果输出。

    解法 ​

    py
    # leetcode submit region begin(Prohibit modification and deletion)
from collections import defaultdict


class Solution:
    def frequencySort(self, s: str) -> str:
        hashmap = defaultdict(int)
        for i in s:
            hashmap[i] += 1

        sorted_desc = dict(sorted(hashmap.items(), key=lambda x: x[1], reverse=True))

        return ''.join(k * v for k, v in sorted_desc.items())



        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N + KlogK)
    • 空间复杂度 O(N + K)