一、题目大意 https://leetcode.cn/problems/sort-characters-by-frequency 给定一个字符串 s ,根据字符出现的 频率 对其进行 降序排序 。一个字符出现的 频率 是它出现在字符串中的次数。 返回 已排序的字符串 。如果有多个答案,返回其中任何一个。 示例1: 输入: s = "tree" 输出: "eert" 解释: 'e'出...
} };classSolution {public:stringfrequencySort(strings) {stringret; map<char,int,cisort>countmap;//获取映射关系for(size_t i =0;i < s.length();++i){if(countmap.find(s.at(i)) !=countmap.end()){ countmap[s.at(i)]++; }else{ countmap[s.at(i)]=1; } }while(countmap.size()...
本题是LeetCode 1636 - 按照频率将数组升序排序 加强版,将数组换成了字符串,使用相同的思路即可通过。 先用一个 map 统计 s 中每个字符的出现次数。 然后对 s 中的字符按照出现次数降序排序,出现次数相同时,按字符升序排序(以保证相同字符在一起)。 最后转成字符串返回即可。 设字符集大小为 C 。 时间复杂...
【Leetcode】451. Sort Characters By Frequency https://leetcode.com/problems/sort-characters-by-frequency/#/description 特殊的排序,对频率排序。 1.常规思路,先遍历一遍,把频数存入map,再把map的entry存入priorityqueue,最后逐一出队即可。代码: public String frequencySort1(String s) { Map<Chara...
解题思路:按照记不得曾经:LeetCode347.Top K Frequent Elements求解即可 classSolution(object):deffrequencySort(self,s):""":type s: str:rtype: str"""res={}foriins:ifiinres:res[i]+=1else:res[i]=1res=sorted(res.items(),key=lambdax:x[1],reverse=True)r=""foriinres:r+=i[0]*i[1...
1publicclassSolution {2publicString frequencySort(String s) {3Map<Character, Integer> map =newHashMap<>();4for(charc : s.toCharArray()) {5if(map.containsKey(c)) {6map.put(c, map.get(c) + 1);7}else{8map.put(c, 1);9}10}11List<Character> [] bucket =newList[s.length() +...
451. Sort Characters By FrequencyMedium Topics Companies Given a string s, sort it in decreasing order based on the frequency of the characters. The frequency of a character is the number of times it appears in the string. Return the sorted string. If there are multiple answers, return any...
classSolution(object):deffrequencySort(self,s):""":type s: str :rtype: str """return''.join(c*tforc,tincollections.Counter(s).most_common())
文章作者:Tyan 博客:noahsnail.com|CSDN|简书 1. Description Sort Characters By Frequency 2. Solution boolcompare(pair<char,int>&a,pair<char,int>&b){returna.second>b.second;}class Solution{public:stringfrequencySort(string s){map<char,int>stat;for(inti=0;i<s.length();i++){stat[s[i]]...
publicStringfrequencySort(Strings) {Map<Character, Integer>map=newHashMap<>();for(int i =0; i < s.length(); i++) {map.put(s.charAt(i),map.getOrDefault(s.charAt(i),0)+1); } PriorityQueue<Map.Entry<Character, Integer>> queue =newPriorityQueue<>((a, b)->b.getValue()-a.get...