Can you solve this real interview question? Sort Characters By Frequency - 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 s
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...
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。 ......
«LeetCode 347. Top K Frequent Elements(出现频率最多的 k 个元素) »LeetCode 75. Sort Colors(按颜色进行排序) posted @2020-03-30 14:36Somnuspoppy阅读(142) 评论(0)编辑 【推荐】100%开源!大型工业跨平台软件C++源码提供,建模,组态!
classSolution {public:stringfrequencySort(strings) { unordered_map<char,int>m;for(constauto&c:s) m[c]++; priority_queue<pair<char,int>, vector<pair<char,int>>, cmp >q;for(constauto&i:m){ q.push(make_pair(i.first, i.second)); }stringres ="";while(!q.empty()){ res.append(...
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...
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() +...
leetcode 451. Sort Characters By Frequency 排序即可,Givenastring,sortitindecreasingorderbasedonthefrequencyofcharacters.Example1:Input:“tree”Output:“eert”Explanation:‘e’appe
public String frequencySort(String s) { Map<Character, Integer> map = new HashMap<>(); 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 = new PriorityQueue<>((a, b)...
https://leetcode.com/problems/sort-characters-by-frequency/?tab=Description 思路1: Priority Queue 先把所有字符的出现次数统计一下---O(n) 然后把 (字符, 出现次数) 做成一个node推入优先队列,按题目要求重载比较规则---O(n log n) 最后依次pop队列元素,构造结果---O(n) Time ...