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...
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...
所以在这种情况下,把数组换成Map / Tree Map是比较好的做法,可以有效避免空间浪费,且具有排序的功能(C++std::map内部是一颗红黑树),最后构造结果的时候从map的最大key开始遍历就可以了: stringfrequencySort(string s){if(s.empty())return""; unordered_map<char,int> cnt;for(autoc : s) { cnt[c]++;...
First Unique Character in a String 参考资料: https://leetcode.com/problems/sort-characters-by-frequency/description/ https://leetcode.com/problems/sort-characters-by-frequency/discuss/93404/c-on-solution-without-sort https://leetcode.com/problems/sort-characters-by-frequency/discuss/93409/concise-c...
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。 ......
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
public String frequencySort(String s) { int[] count = new int[256]; for (char ch : s.toCharArray()) { count[ch]++; } Map<Integer, List<Character>> map = new HashMap<>(); for (int i = 0; i < count.length; i++) { ...
leetcode 451. Sort Characters By Frequency 排序即可,Givenastring,sortitindecreasingorderbasedonthefrequencyofcharacters.Example1:Input:“tree”Output:“eert”Explanation:‘e’appe
string frequencySort(string s) { unordered_map<char,int> freq; vector<string> bucket(s.size()+1, ""); string res; //count frequency of each character for(char c:s) freq[c]++; //put character into frequency bucket for(auto& it:freq) { ...
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)...