Leetcode-451. Sort Characters By Frequency

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

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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.

这个想法就是用一个新的类，保存字符是什么，以及当前字符多少个了，最后对于这个List进行排序即可。Your runtime beats 71.22% of java submissions.

public class Solution {    public String frequencySort(String s) {        Map<Character, Product> map = new HashMap<Character, Product>();        List<Product> list = new ArrayList<Product>();        for(char item : s.toCharArray()){            if(map.containsKey(item)){                Product product = map.get(item);                product.sb.append(item);                product.count ++;            }else{                Product product = new Product();                product.sb.append(item);                product.count = 1;                list.add(product);                map.put(item, product);            }        }        Collections.sort(list, new Comparator<Product>(){            public int compare(Product p1, Product p2){                return p2.count - p1.count;            }        });        StringBuffer sb = new StringBuffer("");        for(Product item : list){            sb.append(item.sb);        }        return sb.toString();    }    class Product{        int count;        StringBuffer sb = new StringBuffer("");    }}