451. Sort Characters By Frequency

Q:

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.

A:

就是放到PriorityQueue里再取出来。

public class Solution {
    public String frequencySort(String s) {
        if(s==null || s.length()==0)
            return "";
        // put to HashMap
        Map<Character,Integer> map = new HashMap();
        for(char ch : s.toCharArray()){
            map.put(ch, (map.containsKey(ch))? map.get(ch)+1: 1);
        }
        PriorityQueue<Obj> maxQueue = new PriorityQueue<>(map.size(), (A,B)->B.count - A.count);
        //from HashMap, put to priorityQueue
        for(char ch : map.keySet()){
            Obj obj = new Obj(ch, map.get(ch));
            maxQueue.add(obj);
        }
        //build result from PriorityQueue
        StringBuffer buf = new StringBuffer();
        while( ! maxQueue.isEmpty()) {
            Obj node = maxQueue.poll();
            for(int i =0;i<node.count;i++)
                buf.append(node.ch);
        }
        return buf.toString();
    }
    class Obj{
        char ch ='a';
        int count = 0;
        public Obj(char c, int cc){
            ch = c;
            count = cc;
        }
    }
}

Errors：

由于PriorityQueue需要的初始capaity 不能小于1。
因此，我们需要开始检查s是否为0.