最长回文子序列的Python程序| DP-12

给定一个序列，找出其中最长的回文子序列的长度。

再举一个例子，如果给定的序列是“BBABCBCAB”，那么输出应该是 7，因为“BABCBAB”是其中最长的回文子序列。 “BBBBB”和“BBCBB”也是给定序列的回文子序列，但不是最长的。
1) 最优子结构：
令 X[0..n-1] 为长度为 n 的输入序列，L(0, n-1) 为 X[0..n-1] 的最长回文子序列的长度。
如果 X 的最后一个字符和第一个字符相同，则 L(0, n-1) = L(1, n-2) + 2。
否则 L(0, n-1) = MAX (L(1, n-1), L(0, n-2))。
以下是处理所有情况的通用递归解决方案。
动态规划解决方案

Python3

# A Dynamic Programming based Python 
# program for LPS problem Returns the length
# of the longest palindromic subsequence in seq
def lps(str):
    n = len(str)
  
    # Create a table to store results of subproblems
    L = [[0 for x in range(n)] for x in range(n)]
  
    # Strings of length 1 are palindrome of length 1
    for i in range(n):
        L[i][i] = 1
  
    # Build the table. Note that the lower 
    # diagonal values of table are
    # useless and not filled in the process. 
    # The values are filled in a
    # manner similar to Matrix Chain 
    # Multiplication DP solution (See
    # https://www.geeksforgeeks.org/
    # dynamic-programming-set-8-matrix-chain-multiplication/
    # cl is length of substring
    for cl in range(2, n + 1):
        for i in range(n-cl + 1):
            j = i + cl-1
            if str[i] == str[j] and cl == 2:
                L[i][j] = 2
            elif str[i] == str[j]:
                L[i][j] = L[i + 1][j-1] + 2
            else:
                L[i][j] = max(L[i][j-1], L[i + 1][j]);
  
    return L[0][n-1]
  
# Driver program to test above functions
seq = "GEEKS FOR GEEKS"
n = len(seq)
print("The length of the LPS is " + str(lps(seq)))
  
# This code is contributed by Bhavya Jain

输出：

The length of the LPS is 7

时间复杂度：- O(n ² )

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