LeetCode 451. Sort Characters By Frequency

原题链接在这里：https://leetcode.com/problems/sort-characters-by-frequency/

题目：

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.

题解：

类似Top K Frequent Elements. 利用Bucket Sort 按照频率把s的char放进bucket里, 再按照频率从大到小，重复频率次append到res中.

Note: When add bucket element to result. Do not forget to check if this element is null or not.

Time Complexity: O(n), n = s.length(). Space: O(n).

AC Java:

public class Solution {
    public String frequencySort(String s) {
        if(s == null || s.length() == 0){
            return s;
        }
        
        HashMap<Character, Integer> freqMap = new HashMap<Character, Integer>();
        for(char c : s.toCharArray()){
            freqMap.put(c, freqMap.getOrDefault(c, 0)+1);
        }
        
        StringBuilder [] bucket = new StringBuilder[s.length()+1];
        for(char c : freqMap.keySet()){
            int freq = freqMap.get(c);
            if(bucket[freq] == null){
                bucket[freq] = new StringBuilder();
            }
            for(int i = 0; i<freq; i++){
                bucket[freq].append(c);
            }
        }
        
        StringBuilder res = new StringBuilder();
        for(int i = bucket.length-1; i>=0; i--){
            if(bucket[i] != null){
                res.append(bucket[i]);
            }
        }
        return res.toString();
    }
}

AC C++:

class Solution {
public:
    string frequencySort(string s) {
        unordered_map<char, int> map;
        for(char c : s){
            map[c]++;
        }   

        int n = s.size();
        vector<vector<char>> bucket(n + 1);
        for(auto [c, f] : map){
            bucket[f].push_back(c);
        }

        string res;
        for(int i = n; i>=0; i--){
            for(char c : bucket[i]){
                res.append(i, c);
            }
        }

        return res;
    }
};

AC Python:

class Solution:
    def frequencySort(self, s: str) -> str:
        map = Counter(s)
        n = len(s)
        bucket = [[] for _ in range(n + 1)]
        for c, freq in map.items():
            bucket[freq].append(c)
        
        res = []
        for freq in range(n, 0, -1):
            for c in bucket[freq]:
                res.append(c * freq)
        
        return "".join(res)

类似Top K Frequent Words, First Unique Character in a String.