451. Sort Characters By Frequency

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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。通过使用unordered_map统计输入字符串中每个字符的出现次数，并利用vector存储不同频次的字符，最终按频次降序输出排序后的字符串。文章提供了详细的代码示例并解释了关键步骤。

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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.

reference: http://blog.youkuaiyun.com/qq508618087/article/details/53125359

首先用unordered_map 统计频率，然后开一个数组，从0-s.size()，其实只用从1-s.size()，因为是频率的范围，刚开始范围弄错了，错了很多次。

细节，string1.append(n,string2),是string1 += n * string2

代码如下：

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