Leetcode 451. Sort Characters By Frequency 按频率对字符排序 解题报告

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();    }}