检查字符串，看看它是否是它的任何排列的回文

Question

我已经创建了一个函数，可以成功地实现这一点（我非常确定），但我担心它的部分效率。我有两个嵌套的for循环，我认为这使得这个算法的最坏情况是O（n^2）。我有什么办法可以改进吗？在

def palindrome(string):
    s = [c.replace(' ', '') for c in string]
    merged = "".join(s)
    srt = sorted(merged)
    dic = {}
    singles = 0

    for i in srt:
        if i not in dic:
            dic[i] = 1
        else:
            # Worried about this second loop having to run for every i in srt
            for key, value in dic.items():
                if key == i:
                    dic[key] = 2 
    for key, value in dic.items():
        if value == 1:
            singles += 1
    if singles > 1:
        return False
    else:
        return True

Answer 1

它测试字符串中字母的任何排列是否可以是回文。在

我建议：

from collections import Counter

def palindrome(string):
    s = string.replace(' ', '')
    return not sum(v % 2 == 1 for k,v in Counter(string)) > 1

此检查奇数字符出现计数的总和不大于1。在