题目地址：https://leetcode.com/problems/count-different-palindromic-subsequences/description/
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.
Asubsequence of a string S is obtained by deleting 0 or more characters from S.
Asequence is palindromic if it is equal to the sequence reversed.
Twosequences 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:
S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba'
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'}.
题目大意
求一个字符串的有多少个回文子序列。
解题方法
记忆化搜索
这个题太难了，我也只是抄了花花酱的答案open in new window，花花有个40分钟的视频，讲得非常清楚，强烈大家看看。
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)
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