String 使用递归方法从字符串中删除一个字符…@Oli Charlesworth:对不起，我没有任何线索：（这与C++或任何特定的编程语言无关。你反问题找到字符串中回文的总数）？字符串中的任何改变都将保持它是回文还是不存在，没有第三个选项，所以找到一个将给你另一个。

---

使用递归方法从字符串中删除一个字符…@Oli Charlesworth:对不起，我没有任何线索：（这与C++或任何特定的编程语言无关。你反问题找到字符串中回文的总数）？字符串中的任何改变都将保持它是回文还是不存在，没有第三个选项，所以找到一个将给你另一个。

numWays(str): return 0 if str is empty
              return memo[str] if it exists
              memo[str] = numWays(str - firstChar) +
                          numWays(str - secondChar) +
                          ... +
                          1 if str is not a palindrome
              return memo[str]

#include <stdio.h>
#include <string.h>

char string[] = "AbbA";
int is_palindrome (char *str, size_t len, size_t mask);

int main(void)
{
size_t len,mask, count;

len = strlen(string);

count =0;
for (mask = 1; mask < (1ul <<len) -1; mask++) {
        if ( is_palindrome (string, len, mask)) continue;
        count++;
        }

fprintf(stderr, "Len:=%u, Count=%u \n"
        , (unsigned) len , (unsigned) count );

return 0;
}

int is_palindrome (char *str, size_t len, size_t mask)
{
size_t l,r,pop;

for (pop=l=0, r = len -1; l < r; ) {
        if ( mask & (1u <<l)) { l++; continue; }
        if ( mask & (1u <<r)) { r--; continue; }
        if ( str[l] == str[r] ) return 1;
        l++,r--; pop++;
        }
return (pop <1) ? 1: 0;
}

import Data.List

listNonPalindromes string = 
  filter (isNotPalindrome) (subsequences string)
    where isNotPalindrome str
            | fst substr == snd substr  = False
            | otherwise                 = True
           where substr = let a = splitAt (div (length str) 2) str
                          in (reverse (fst a), if even (length str) 
                                                  then snd a 
                                                  else drop 1 (snd a))

howManyNonPalindromes string = length $ listNonPalindromes string

# Count the palindromic subsequences of s
def pcount(s):
  length = len(s)
  p0 = [0] * (length + 1)
  p1 = [1] * length
  for l in range(1, length):
    for i in range(length - l):
      p0[i] = p1[i]
      if s[i] == s[i + l]:
        p1[i] += p1[i+1] + 1
      else:
        p1[i] += p1[i+1] - p0[i+1]
  # The "+ 1" is to account for the empty sequence, which is a palindrome.
  return p1[0] + 1

# Count the non-palindromic subsequences of s
def npcount(s):
  return 2**len(s) - pcount(s)