题目

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

本题思路很简单，是个典型的hashmap+优先队列的思想，总共只需要遍历三遍；分为以下三步：

• step1:遍历一遍字符串，同时存储字符并统计出现的次数
• step2:遍历哈希表，利用优先队列排序
• step3:遍历优先队列，输出结果

具体流程见代码：

代码

class Solution {
public:
    string frequencySort(string s) {
        unordered_map<char,int> mp;
        priority_queue<pair<int,char>> pq;
        for(auto item:s)
        {
            mp[item]++;
        }
        for(auto it = mp.begin();it!=mp.end();it++)
        {
            pq.push(make_pair(it->second,it->first));
        }
        string res ="";
        while(!pq.empty())
        {
            for(int i=0;i<pq.top().first;i++)
            {
                res+=pq.top().second;
            }
            pq.pop();
        }
        return res;
    }
};

最后

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