[leetcode] 451. Sort Characters By Frequency 解题报告

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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