1 解题思想

题目要求给一个字符串，要求将字符串中的字符，按照字符出现的频率进行排序，从大到小的输出（某个字符出现了多少就输出多少次，只是重组）

解题方法：
1、统计
2、使用TreeMap，将相同频率的字符存在一起，然后按照key=频率的方式放入TreeMap
3、遍历TreeMap输出

2 原题

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.

3 AC解

public class Solution {
    public String frequencySort(String s) {
        int frequency[] = new int[256];
        for(char c:s.toCharArray()){
            frequency[c] ++;
        }
        // 统计频率，使用Treemap排序,这里白板写的代码，直接将频率取反，这样方便
        TreeMap<Integer,String> map= new TreeMap<Integer,String>();
        for(int i=0;i<256;i++){
            if(frequency[i] == 0) continue;
            StringBuilder sb = new StringBuilder(map.getOrDefault(-frequency[i],""));
            for(int j=0;j<frequency[i];j++)
                sb.append((char)i);
            map.put(-frequency[i],sb.toString());
        }
        StringBuilder sb = new StringBuilder();
        Iterator it = map.keySet().iterator();
        while(it.hasNext()){
            Integer key = (Integer)it.next();
            sb.append(map.get(key));
        }
        return sb.toString();
    }
}

最后

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