# 451. Sort Characters By Frequency

Medium
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.

### 解題

func frequencySort(s string) string {
count := make(map[byte]int)
bytes := []byte(s)
for _, c := range bytes {
count[c]++
}
sort.Slice(bytes, func(i, j int) bool {
if count[bytes[i]] == count[bytes[j]] {
return bytes[i] > bytes[j]
}
return count[bytes[i]] > count[bytes[j]]
})
return string(bytes)
}