这是我参与8月更文挑战的第14天，活动详情查看：8月更文挑战
描述
Given a string s, return true if s is a good string, or false otherwise.
A string s is good if all the characters that appear in s have the same number of occurrences (i.e., the same frequency).
Example 1:
Input: s = "abacbc"
Output: true
Explanation: The characters that appear in s are 'a', 'b', and 'c'. All characters occur 2 times in s.
Example 2:
Input: s = "aaabb"
Output: false
Explanation: The characters that appear in s are 'a' and 'b'.
'a' occurs 3 times while 'b' occurs 2 times, which is not the same number of times.
Note:
1 <= s.length <= 1000
s consists of lowercase English letters.
解析
根据题意，就是给出一个字符串 s ，判断这个字符串 s 是不是 good 。题目中给出了 good 字符串的定义，就是字符串 s 中的每个出现的字符的个数相等。思路很简单：
- 初始化一个计数器 count
- 遍历 s 中的每个字符，对他们出现的数量进行计数
- 判断 count.values() 列表中的数字在去重之后长度是否为 1 ，如果为 1 则表示 s 是 good 字符串，否则不是 good 字符串
解答
class Solution(object):
def areOccurrencesEqual(self, s):
"""
:type s: str
:rtype: bool
"""
count = {}
for c in s:
if c not in count:
count[c] = 1
else:
count[c] += 1
return len(set(count.values()))==1
运行结果
Runtime: 24 ms, faster than 80.22% of Python online submissions for Check if All Characters Have Equal Number of Occurrences.
Memory Usage: 13.6 MB, less than 13.55% of Python online submissions for Check if All Characters Have Equal Number of Occurrences.
解析
本质上这个题目考的就是计数器的使用，所以可以借用 python 的内置函数 collections.Counter() 得到对 s 中的字符的技术结果 c ，然后对 c.values() 去重判断是否长度为 1 ，和上面思路一样。
解答
class Solution(object):
def areOccurrencesEqual(self, s):
"""
:type s: str
:rtype: bool
"""
c = collections.Counter(s)
return len(set(c.values()))==1
运行结果
Runtime: 32 ms, faster than 38.10% of Python online submissions for Check if All Characters Have Equal Number of Occurrences.
Memory Usage: 13.5 MB, less than 70.33% of Python online submissions for Check if All Characters Have Equal Number of Occurrences.
原题链接：leetcode.com/problems/ch…
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