451. Sort Characters By Frequency

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# 題目敘述

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.

# 解題思路

# Solution

import java.util.HashMap;

import java.util.Map;

import java.util.PriorityQueue;

class Solution {

    public String frequencySort(String s) {

        Map<Character, Integer> map = new HashMap<>();

        for (char c : s.toCharArray()) {

            map.put(c, map.getOrDefault(c, 0) + 1);

        }

        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>(

            (a, b) -> b.getValue() - a.getValue()

        );

        pq.addAll(map.entrySet());

        StringBuilder ans = new StringBuilder();

        while (!pq.isEmpty()) {

            Map.Entry<Character, Integer> entry = pq.poll();

            ans.append(String.valueOf(entry.getKey()).repeat(entry.getValue()));

        }

        return ans.toString();

    }

}