# 730. Count Different Palindromic Subsequences 统计不同回文子序列

@TOC

## # 题目描述

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'}.

## # 解题方法

### # 记忆化搜索

``````class Solution:
def countPalindromicSubsequences(self, S):
"""
:type S: str
:rtype: int
"""
def count(S, i, j):
if i > j: return 0
if i == j: return 1
if self.m_[i][j]:
return self.m_[i][j]
if S[i] == S[j]:
ans = count(S, i + 1, j - 1) * 2
l = i + 1
r = j - 1
while l <= r and S[l] != S[i]: l += 1
while l <= r and S[r] != S[i]: r -= 1
if l > r: ans += 2
elif l == r: ans += 1
else: ans -= count(S, l + 1, r - 1)
else:
ans = count(S, i + 1, j) + count(S, i, j - 1) - count(S, i + 1, j - 1)

self.m_[i][j] = ans % (10 ** 9 + 7)
return self.m_[i][j]

n = len(S)
self.m_ = [[None for _ in range(n)] for _ in range(n)]
return count(S, 0, n - 1)
``````

## # 日期

2018 年 11 月 17 日 —— 美妙的周末，美丽的天气