题:https://leetcode.com/problems/sort-characters-by-frequency/description/
题目
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.
题目大意
按字符串 出现频率高低 排列字符串。
思路
桶排序,一般按 频率排序等,都可以考虑桶排序。
- 统计 string,计算每个 char出现的频率。
- 桶排序,生成 bucketsList 数组 (长为 string的长度 + 1),List[i] 为出现了 第i次的字符的list。
- 然后从高到低,遍历 bucketsList ,进行输出。
class Solution {
public String frequencySort(String s) {
Map<Character,Integer> frequencyMap = new HashMap<>();
for(int i =0 ;i<s.length();i++)
frequencyMap.put(s.charAt(i),frequencyMap.getOrDefault(s.charAt(i),0)+1);
List<Character>[] bucketsList = new List[s.length()+1];
for(char ch:frequencyMap.keySet()){
if(bucketsList[frequencyMap.get(ch)]==null)
bucketsList[frequencyMap.get(ch)] = new ArrayList<>();
bucketsList[frequencyMap.get(ch)].add(ch);
}
StringBuilder res = new StringBuilder();
for(int i = bucketsList.length-1;i>=1;i--)
if(bucketsList[i]!=null)
for(char ch:bucketsList[i])
for(int j = 0;j < i;j++)
res.append(ch);
return res.toString();
}
}