[leetcode] 451. Sort Characters By Frequency 解题报告
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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
思路:一个直观的思路是先统计每个字符的数量,然后将这些字符连同频率放入一个优先队列中,再取出来即可.这种时间复杂度为O(n) + O(m log m),其中n为字符串长度,m为不同字符的个数,在最坏情况下时间复杂度为O(n log n),即所有字符都不一样.其实我们还有一种可以优化的方法,在统计完字符频率之后利用类似与计数排序的方法,开一个n+1长度大小的数组,将不同的频率字符放到频率的索引处.然后从高到低取得所有字符串.这种方法的好处是在最环情况下依然可以保证时间复杂度为O(n).
代码如下:
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; }};
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