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...
classSolution{public:stringfrequencySort(strings){intcounts[256]={0};for(charch:s)++counts[ch];sort(s.begin(),s.end(),[&](chara,charb){returncounts[a]>counts[b]||(counts[a]==counts[b]&&a<b);});returns;}}; 希望大家关注一下我,我会每天都发一篇文章来感谢大家,也欢迎关注我的微信...
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....
LeetCode #451 - Sort Characters By Frequency 题目描述: Given a string, sort it in decreasing order based on the frequency of characters. Example 1: Example 2: Example 3: 对于一个字符串,按照字符的频率排序。将字符和字符的频率组成pair,然后按照频率进行排序,进而构造新的字符串。 ......
classSolution {public:stringfrequencySort(strings) {stringres =""; vector<string> v(s.size() +1,""); unordered_map<char,int>m;for(charc : s) ++m[c];for(auto&a : m) { v[a.second].append(a.second, a.first); }for(inti = s.size(); i >0; --i) {if(!v[i].empty()...
package leetcode import ( "sort" ) func frequencySort(s string) string { if s == "" { return "" } sMap := map[byte]int{} cMap := map[int][]byte{} sb := []byte(s) for _, b := range sb { sMap[b]++ } for key, value := range sMap { cMap[value] = append(cMap[va...
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]]...