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[力扣刷题攻略] Re#xff1a;从零开始的力扣刷题生活
力扣题1#xff1a;451. 根据字符出现频率排序
解题思想#xff1a;统计字符出现的个数#xff0c;进行排序即可 class Solution(object):def frequencySort(self, s)::type s: str:…力扣题-12.1
[力扣刷题攻略] Re从零开始的力扣刷题生活
力扣题1451. 根据字符出现频率排序
解题思想统计字符出现的个数进行排序即可 class Solution(object):def frequencySort(self, s)::type s: str:rtype: strchar_count {}for i in range(len(s)):if s[i] in char_count:char_count[s[i]] 1else:char_count[s[i]] 1sorted_char_count OrderedDict(sorted(char_count.items(), keylambda item: item[1], reverseTrue))result for char, count in sorted_char_count.items():result result char * countreturn resultclass Solution {
public:string frequencySort(string s) {std::unordered_mapchar, int char_count;for (char c : s) {if (char_count.find(c) ! char_count.end()) {char_count[c] 1;} else {char_count[c] 1;}}std::vectorstd::pairchar, int char_freq_vec(char_count.begin(), char_count.end());std::sort(char_freq_vec.begin(), char_freq_vec.end(),[](const auto a, const auto b) {return a.second b.second;});std::string result;for (const auto p : char_freq_vec) {result std::string(p.second, p.first);}return result;}
};