[leetcode] 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.

这道题是按字符出现频率排序字符串，题目难度为Medium。

首先统计每个字符出现的次数，后续处理思路有很多种，这里借助桶排序的方法来方便最终字符串的构筑。具体代码：

class Solution {public:    string frequencySort(string s) {        vector<int> freq(256, 0);        vector<string> bucket(s.size()+1, "");        string ret = "";                for(auto ch:s) ++freq[ch];        for(int i=0; i<256; ++i)            bucket[freq[i]] += string(freq[i], i);        for(int i=bucket.size()-1; i>0; --i)            ret += bucket[i];                    return ret;    }};

统计完每个字符出现的次数后，将次数相同的字符放在同一个桶中，最终从大到小遍历所有桶即可得出排序后的字符串，时间复杂度为O(n)。

这道题和第347题非常相似，大家可以顺便看下第347题（传送门）。同理这里也可以用堆来进行处理，依据字符出现次数入堆，然后依次从堆中取出各字符即可得出最终结果。由于需要建堆，时间复杂度为O(nlogn)。具体代码：

class Solution {public:    string frequencySort(string s) {        vector<int> freq(256, 0);        priority_queue<pair<int, char>> heap;        string ret = "";                for(auto ch:s) ++freq[ch];        for(int i=0; i<256; ++i)            if(freq[i]) heap.push(make_pair(freq[i], i));        while(!heap.empty()) {            pair<int, char> p = heap.top();            ret += string(p.first, p.second);            heap.pop();        }                return ret;    }};