题目:
Given a string, sort it in decreasing order based on the frequency of characters.
Example 1:
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9Input: "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:
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9Input: "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:
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9Input: "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:
这道题目如果采用基于比较的排序,则思路也很简单,不过时间复杂度最少需要O(nlogn)。但是如果采用计数排序,则可以将时间复杂度降低到O(n)。所以这道题目的考点我感觉其实是计数排序,请参考下面的代码。
代码:
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19class Solution { public: string frequencySort(string s) { int len = s.size(); unordered_map<char, int> hash; for(auto ch : s) { ++hash[ch]; } vector<string> vec(len + 1, ""); for(auto val : hash) { vec[val.second].append(val.second, val.first); } string ret; for(int i = len; i >= 0; --i) { ret += vec[i]; } return ret; } };
最后
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