leetcode1370

 1 class Solution:
 2     def sortString(self, s: str) -> str:
 3         n = len(s)
 4         dic = {}
 5         for i in range(n):
 6             if s[i] not in dic:
 7                 dic[s[i]] = 1
 8             else:
 9                 dic[s[i]] += 1
10         sorted_dic1 = sorted(dic.items(),key=lambda x:(x[0]))
11         sorted_dic2 = sorted(dic.items(),key=lambda x:(x[0]),reverse=True)
12         count = 0
13         result = ''
14         while count < n:
15             for d in sorted_dic1:
16                 key = d[0]
17                 if dic[key] > 0:
18                     dic[key] -= 1
19                     result += key
20                     count += 1
21             for d in sorted_dic2:
22                 key = d[0]
23                 if dic[key] > 0:
24                     dic[key] -= 1
25                     result += key
26                     count += 1
27         return result

Algorithm thinking: hash.

Use a dictionary record the number of times each character appears on the dictionary sort twice.

According to the sort of positive sequence of characters, the characters sorted in reverse order.

A "snake" cycle, the first positive sequence traversing the dictionary, each character removed from a splice to result string; then traverse the dictionary in reverse order, each time removed from a character string splicing to result.

When the number n of the splice, a description of all characters have been spliced ​​to the result string and returns the result string.