451. Sort Characters By Frequency

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.

笔记：有时候链表数组不失为一种好的数据结构，并且，HASHMAP也作为一个快速查找方案值得考虑。

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

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