451. Sort Characters By Frequency

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.

首先用hashmap扫一遍string，统计频率。按频率排序，将hashmap的entry存入max heap，再一个个poll出来，append到StringBuilder

时间：O(NlogN)，空间：O(N)

class Solution {
    public String frequencySort(String s) {
        HashMap<Character, Integer> map = new HashMap<>();
        PriorityQueue<Map.Entry<Character, Integer>> maxHeap = new PriorityQueue<>((a, b) -> (b.getValue() - a.getValue()));
        
        for(char c : s.toCharArray()) {
            map.put(c, map.getOrDefault(c, 0) + 1);
        }
        
        for(Map.Entry<Character, Integer> entry : map.entrySet()) {
            maxHeap.offer(entry);
        }
        
        StringBuilder sb = new StringBuilder();
        while(!maxHeap.isEmpty()) {
            Map.Entry<Character, Integer> tmp = maxHeap.poll();
            for(int j = 0; j < tmp.getValue(); j++)
                sb.append(tmp.getKey());
        }
        return sb.toString();
    }
}

class Solution {
    public String frequencySort(String s) {
        Map<Character, Integer> map = new HashMap<>();
        for(int i = 0; i < s.length(); i++) {
            map.put(s.charAt(i), map.getOrDefault(s.charAt(i), 0) + 1);
        }
        
        PriorityQueue<Map.Entry<Character, Integer>> maxHeap = new PriorityQueue<>((a, b) -> b.getValue() - a.getValue());
        maxHeap.addAll(map.entrySet());
        
        StringBuilder sb = new StringBuilder();
        while(!maxHeap.isEmpty()) {
            Map.Entry<Character, Integer> entry = maxHeap.poll();
            for(int i = 0; i < entry.getValue(); i++)
                sb.append(entry.getKey());
        }
        return sb.toString();
    }
}