第一次代码如下: classcisort{public:booloperator() (pair<char,int> p1,pair<char,int> p2)const{returnp1.second >p2.second; } };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(countm...
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]++;...
451. Sort Characters By Frequency # 题目# 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'. ...
func frequencySort(s string) string { // chToCnt[ch] 表示 s 中 ch 的出现次数 chToCnt := make(map[rune]int) for _, ch := range s { chToCnt[ch] += 1 } // 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) chs := ([]rune)(s)...
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
leetcode 451. Sort Characters By Frequency 排序即可 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....
LeetCode-Sort Characters By Frequency Description: 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....
Given a string, sort it in decreasing(降序) order based on the frequency(频率) 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 ...
0451-Sort-Characters-By-Frequency/cpp-0451 CMakeLists.txt main.cpp 0454-4Sum-II 0455-Assign-Cookies 0470-Implement-Rand10-Using-Rand7 0473-Matchsticks-to-Square 0474-Ones-and-Zeroes 0478-Generate-Random-Point-in-a-Circle 0485-Max-Consecutive-Ones 0490-The-Maze 0494-Target-Sum 0497-Random-Poin...