Given a string, sort it in decreasing order based on the frequency of characters.
Example 1:
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:
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:
Output:
“bbAa”
Explanation:
“bbaA” is also a valid answer, but “Aabb” is incorrect.
Note that ‘A’ and ‘a’ are treated as two different characters.
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;
}
};