LeetCode 451. Sort Characters By Frequency （根据字符出现频率排序）

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.

 

 

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 题目标签：Hash Table | Heap

　　题目给了我们一个string s， 让我们把s 按照每一个char 的出现频率从大到小排序。

　　首先遍历string s，把每一个char 当作 key， 出现次数当作 value 存入 map；

　　接着利用priorityQueue，按照 value 从大到小的 顺序 排列，把map 的data 存入 pq；

　　遍历 pq，把每一个char 按照 它的value 组成对应长度 string 存入 res。

 

 

Java Solution:

Runtime beats 40.06% 

完成日期：06/23/2017

关键词：HashMap; Priority Queue

关键点：把Map 存入 pq

class Solution 
{
    public String frequencySort(String s) 
    {
        HashMap<Character, Integer> map = new HashMap<>();
        StringBuilder res = new StringBuilder();
        
        // store s char and frequency into map
        for(char c: s.toCharArray())
            map.put(c, map.getOrDefault(c, 0) + 1);
        
        // set up priorityQueue in value descending order
        PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>(
            new Comparator<Map.Entry<Character, Integer>>()
            {
                public int compare(Map.Entry<Character, Integer> a, Map.Entry<Character, Integer> b)
                {
                    return b.getValue() - a.getValue();
                }
            }
        );
        
        // add map into priorityQueue
        pq.addAll(map.entrySet());
        
        // iterate pg, add each char with value times into res 
        while(!pq.isEmpty())
        {
            Map.Entry<Character, Integer> entry = pq.poll();
            
            for(int i=0; i<(int) entry.getValue(); i++)
                res.append(entry.getKey());
        }
        
        return res.toString();
    }
}

参考资料：

https://discuss.leetcode.com/topic/66024/java-o-n-bucket-sort-solution-o-nlogn-priorityqueue-solution-easy-to-understand

 

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