## 题目描述：

LeetCode 730. Count Different Palindromic Subsequences

Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo `10^9 + 7`.

A subsequence of a string S is obtained by deleting 0 or more characters from S.

A sequence is palindromic if it is equal to the sequence reversed.

Two sequences `A_1, A_2, ...` and `B_1, B_2, ...` are different if there is some `i` for which `A_i != B_i`.

Example 1:

```Input:
S = 'bccb'
Output: 6
Explanation:
The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'.
Note that 'bcb' is counted only once, even though it occurs twice.
```

Example 2:

```Input:
Output: 104860361
Explanation:
There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.
```

Note:

• The length of `S` will be in the range `[1, 1000]`.
• Each character `S[i]` will be in the set `{'a', 'b', 'c', 'd'}`.

## Python代码：

``````class Solution(object):
def countPalindromicSubsequences(self, S):
"""
:type S: str
:rtype: int
"""
size = len(S)
next = [{k : -1 for k in 'abcd'} for x in range(size + 1)]
prev = [{k : -1 for k in 'abcd'} for x in range(size + 1)]
for x in range(size):
for k in 'abcd':
if S[x] == k: prev[x][k] = x
else: prev[x][k] = prev[x - 1][k]

for x in range(size - 1, -1, -1):
for k in 'abcd':
if S[x] == k: next[x][k] = x
else: next[x][k] = next[x + 1][k]

dmap = [[0] * (size + 1) for x in range(size + 1)]

def solve(i, j):
if i > j: return 0
if dmap[i][j]: return dmap[i][j]
ans = 0
for k in 'abcd':
ii, jj = next[i][k], prev[j][k]
if ii < 0: continue
if ii < jj: ans += 1
if ii <= j: ans += solve(ii + 1, jj - 1) + 1
dmap[i][j] = ans % (10 ** 9 + 7)
return dmap[i][j]

return solve(0, size - 1)
``````

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