leetcode 451. Sort Characters By Frequency

最新推荐文章于 2022-07-30 00:29:18 发布

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最新推荐文章于 2022-07-30 00:29:18 发布

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本文介绍了一种基于字符出现频率进行字符串排序的算法实现。通过使用C++标准模板库中的map和vector，文章详细展示了如何统计字符串中每个字符的出现次数，并按频率从高到低对字符进行排序。

451. 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.

1、

static bool compare(pair<char,int>a,pair<char,int>b)
{
return a.second > b.second;
}

sort(pairArray.begin(), pairArray.end(), compare);

这种自定义排序一定要会。

2、这个题我本来用 multimap<int, char> 来转存 map<char, int> chars的，思路是一样的。但是会MLE。非常费解！

static bool compare(pair<char,int>a,pair<char,int>b) 
{  
    return a.second > b.second;  
}

class Solution {
public:
    string frequencySort(string s) 
    {
        map<char, int> chars;
        for (int i = 0; i < s.size(); i++)
        {
            chars[s[i]]++;
        }
        vector<pair<char,int>> pairArray;
        for (auto it = chars.begin(); it != chars.end(); it++)
        {
            pairArray.push_back(make_pair(it->first, it->second));
        } 
        string ret = "";
        sort(pairArray.begin(), pairArray.end(), compare);  
        for (pair<char,int> pair : pairArray) 
        {  
            ret.append(pair.second, pair.first);  
        }  
        return ret;
    }
};