[leetcode]451. Sort Characters By Frequency-优快云博客

本文链接：https://blog.youkuaiyun.com/xiaocong1990/article/details/54233105

本文介绍了一种基于字符出现频率对字符串进行排序的方法。通过使用哈希表统计字符频率，并利用数组辅助存储，最终实现字符串中字符按频率从高到低排序。文章提供了详细的算法步骤及代码实现。

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题目链接 :https://leetcode.com/problems/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.

解题思路：先统计每个字符出现的频率，然后新开一个长度为字符串长度再加1的数组，将不同频率的字符放到数组索引与频率相同处，然后从高到低取得各个字符，时间复杂度为O(n)。

class Solution{
public:
    string frequencySort(string s)
    {
        int len=s.size();
        unordered_map<char,int> hash;
        vector<string> vec(len+1,"");
        for(auto ch:s) hash[ch]++;
        for(auto val:hash) vec[val.second].append(val.second,val.first);
        string ans;
        for(int i=len;i>0;i--)
            ans+=vec[i];
        return ans;
    }
};