分散回文 - 如何通过字典解析以找出形成回文的子串组合

Question

我在面试时遇到了这个 HC 问题。我能够想出我称之为蛮力方法的方法。这是问题陈述：

Find all the scatter palindromes in a given string, "aabb". The substrings can be scattered but rearranged to form a palindrome. example: a, aa, aab, aabb, a, abb, b, bb, bba and b are the substrings that satisfy this criteria.

我的逻辑：

divide the str into substrings
counter = 0
if the len(substr) is even:
  and substr == reverse(substr)
  increment counter
else:
  store all the number of occurrences of each str of substr in a dict
  process dict somehow to figure out if it can be arranged into a palindrome
  ###### this is where I ran out of ideas ######

我的代码：

class Solution:
def countSubstrings(self, s: str) -> int:
    n = len(s)
    c=0
    for i in range(0,n-1): #i=0
        print('i:',i)
        for j in range(i+1,n+1): #j=1
            print('j',j)
            temp = s[i:j]
            if len(temp) == 1:
                c+=1
            # if the len of substring is even,
            # check if the reverse of the string is same as the string
            elif(len(temp)%2 == 0):
                if (temp == temp[::-1]):
                    c+=1
                    print("c",c)
            else:
                # create a dict to check how many times
                # each value has occurred
                d = {}
                for l in range(len(temp)):
                    if temp[l] in d:
                        d[temp[l]] = d[temp[l]]+1
                    else:
                        d[temp[l]] = 1
                print(d)

    return c


op = Solution()
op.countSubstrings('aabb')

到现在为止，很明显我是一个初学者。我确信有更好、更复杂的方法来解决这个问题。我的一些代码改编自 visleck 的逻辑 here，我无法理解它的后半部分。如果有人能解释一下，那就太好了。

Answer 1

作为部分答案，判断一个字符串是否为分散回文很简单：如果出现奇数次的字母个数最多为1，则为分散回文。否则不是。

可以这样实现：

from collections import Counter

def scattered_palindrome(s):
    counts = Counter(s)
    return len([count for count in counts.values() if count % 2 == 1]) <= 1

例如，

>>> scattered_palindrome('abb')
True
>>> scattered_palindrome('abbb')
False

请注意，在任何阶段都没有必要将字符串与其反向进行比较。另外请注意，我使用了 Counter 对象来跟踪字母计数。这是创建类似字典的字母计数集合的简化方法。

分散回文 - 如何通过字典解析以找出形成回文的子串组合

Scatter palindrome - How to parse through a dictionary to figure out combinations of substrings that form a palindrome

python

algorithm

data-structures

dynamic-programming

palindrome