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.

题目标签:Hash Table | Heap

  题目给了我们一个string s, 让我们把s 按照每一个char 的出现频率从大到小排序。

  首先遍历string s,把每一个char 当作 key, 出现次数当作 value 存入 map;

  接着利用priorityQueue,按照 value 从大到小的 顺序 排列,把map 的data 存入 pq;

  遍历 pq,把每一个char 按照 它的value 组成对应长度 string 存入 res。

Java Solution:

Runtime beats 40.06%

完成日期:06/23/2017

关键词:HashMap; Priority Queue

关键点:把Map 存入 pq

 class Solution
{
public String frequencySort(String s)
{
HashMap<Character, Integer> map = new HashMap<>();
StringBuilder res = new StringBuilder(); // store s char and frequency into map
for(char c: s.toCharArray())
map.put(c, map.getOrDefault(c, 0) + 1); // set up priorityQueue in value descending order
PriorityQueue<Map.Entry<Character, Integer>> pq = new PriorityQueue<>(
new Comparator<Map.Entry<Character, Integer>>()
{
public int compare(Map.Entry<Character, Integer> a, Map.Entry<Character, Integer> b)
{
return b.getValue() - a.getValue();
}
}
); // add map into priorityQueue
pq.addAll(map.entrySet()); // iterate pg, add each char with value times into res
while(!pq.isEmpty())
{
Map.Entry<Character, Integer> entry = pq.poll(); for(int i=0; i<(int) entry.getValue(); i++)
res.append(entry.getKey());
} return res.toString();
}
}

参考资料:

https://discuss.leetcode.com/topic/66024/java-o-n-bucket-sort-solution-o-nlogn-priorityqueue-solution-easy-to-understand

LeetCode 题目列表 - LeetCode Questions List

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