Leetcode 451 Sort Characters By Frequency

最新推荐文章于 2019-07-20 14:12:39 发布

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本文介绍了一种基于字符出现频率进行排序的算法实现。通过统计输入字符串中每个字符的出现次数，并按频率从高到低排序输出。文章提供了一个具体的Java实现示例，包括使用HashMap进行字符计数和构建List数组进行排序。

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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.

思路比较简单，先统计再排序。

注意有可能出现相同频率的，所以说用一个list组成的array来进行排序。

public class Solution {
    public String frequencySort(String s) {
        if(s == null ){
            return null;
        }
        if(s.length() == 0){
            return "";
        }
        Map<Character, Integer> map = new HashMap<>();
        int max = 0;
        for(int i = 0; i < s.length(); i++){
            char c = s.charAt(i);
            if(map.containsKey(c)){
                map.put(c, map.get(c) + 1);
            }else{
                map.put(c, 1);
            }
            
            max = Math.max(max, map.get(c)   );
        }
          List<Character>[] list = new List[max + 1];
        for (Character c : map.keySet()) {
        int count = map.get(c);
            if(list[count] == null){
                list[count] = new ArrayList();
            }
            list[count].add(c);
        }
        
         StringBuilder sb = new StringBuilder();
        for(int i = list.length - 1; i >= 0; i--){
            if(list[i] != null){
                for(Character c : list[i]){
                    for(int j = 0; j < i; j++){
                        sb.append(c);
                    }
                }
            }
        }
        return sb.toString();
    }
}