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();
    }
}