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: S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba' Output: 104860361 Explanation: There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7.
Note:
- The length of
Swill be in the range[1, 1000]. - Each character
S[i]will be in the set{'a', 'b', 'c', 'd'}.
题目大意：
求字符串S的回文子序列的个数
解题思路：
记忆化搜索（Search + Memoization）
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)
本文链接：http://bookshadow.com/weblog/2017/11/20/leetcode-count-different-palindromic-subsequences/
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