最长回文序列| DP-12

给定一个序列，找到其中最长回文子序列的长度。

作为另一个示例，如果给定序列为“ BBABCBCAB”，则输出应为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序列，但不是最长的。
这个问题的幼稚解决方案是生成给定序列的所有子序列，并找到最长的回文子序列。该解决方案在时间复杂度方面是指数级的。让我们看看这个问题如何同时具有动态编程(DP)问题的两个重要属性，以及如何使用动态编程有效地解决。
1)最佳子结构：
令X [0..n-1]为长度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。

// Every single character is a palindrome of length 1
L(i, i) = 1 for all indexes i in given sequence

// IF first and last characters are not same
If (X[i] != X[j])  L(i, j) =  max{L(i + 1, j),L(i, j - 1)} 

// If there are only 2 characters and both are same
Else if (j == i + 1) L(i, j) = 2  

// If there are more than two characters, and first and last 
// characters are same
Else L(i, j) =  L(i + 1, j - 1) + 2 

2)重叠子问题
以下是LPS问题的简单递归实现。该实现仅遵循上述递归结构。

C++

// C++ program of above approach
#include
using namespace std;
 
// A utility function to get max of two integers
int max (int x, int y) { return (x > y)? x : y; }
 
// Returns the length of the longest palindromic subsequence in seq
int lps(char *seq, int i, int j)
{
// Base Case 1: If there is only 1 character
if (i == j)
    return 1;
 
// Base Case 2: If there are only 2
// characters and both are same
if (seq[i] == seq[j] && i + 1 == j)
    return 2;
 
// If the first and last characters match
if (seq[i] == seq[j])
    return lps (seq, i+1, j-1) + 2;
 
// If the first and last characters do not match
return max( lps(seq, i, j-1), lps(seq, i+1, j) );
}
 
/* Driver program to test above functions */
int main()
{
    char seq[] = "GEEKSFORGEEKS";
    int n = strlen(seq);
    cout << "The length of the LPS is "
         << lps(seq, 0, n-1);
    return 0;
}
 
// This code is contributed
// by Akanksha Rai

C

// C program of above approach
#include
#include
 
// A utility function to get max of two integers
int max (int x, int y) { return (x > y)? x : y; }
 
// Returns the length of the longest palindromic subsequence in seq
int lps(char *seq, int i, int j)
{
   // Base Case 1: If there is only 1 character
   if (i == j)
     return 1;
 
   // Base Case 2: If there are only 2 characters and both are same
   if (seq[i] == seq[j] && i + 1 == j)
     return 2;
 
   // If the first and last characters match
   if (seq[i] == seq[j])
      return lps (seq, i+1, j-1) + 2;
 
   // If the first and last characters do not match
   return max( lps(seq, i, j-1), lps(seq, i+1, j) );
}
 
/* Driver program to test above functions */
int main()
{
    char seq[] = "GEEKSFORGEEKS";
    int n = strlen(seq);
    printf ("The length of the LPS is %d", lps(seq, 0, n-1));
    getchar();
    return 0;
}

Java

//Java program of above approach
 
class GFG {
 
    // A utility function to get max of two integers
    static int max(int x, int y) {
        return (x > y) ? x : y;
    }
    // Returns the length of the longest palindromic subsequence in seq
 
    static int lps(char seq[], int i, int j) {
    // Base Case 1: If there is only 1 character
        if (i == j) {
            return 1;
        }
 
    // Base Case 2: If there are only 2 characters and both are same
        if (seq[i] == seq[j] && i + 1 == j) {
            return 2;
        }
 
    // If the first and last characters match
        if (seq[i] == seq[j]) {
            return lps(seq, i + 1, j - 1) + 2;
        }
 
    // If the first and last characters do not match
        return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
    }
 
 
    /* Driver program to test above function */
    public static void main(String[] args) {
        String seq = "GEEKSFORGEEKS";
        int n = seq.length();
        System.out.printf("The length of the LPS is %d", lps(seq.toCharArray(), 0, n - 1));
 
    }
}

Python3

# Python 3 program of above approach
 
# A utility function to get max
# of two egers
def max(x, y):
    if(x > y):
        return x
    return y
     
# Returns the length of the longest
# palindromic subsequence in seq
def lps(seq, i, j):
     
    # Base Case 1: If there is
    # only 1 character
    if (i == j):
        return 1
 
    # Base Case 2: If there are only 2
    # characters and both are same
    if (seq[i] == seq[j] and i + 1 == j):
        return 2
     
    # If the first and last characters match
    if (seq[i] == seq[j]):
        return lps(seq, i + 1, j - 1) + 2
 
    # If the first and last characters
    # do not match
    return max(lps(seq, i, j - 1),
               lps(seq, i + 1, j))
 
# Driver Code
if __name__ == '__main__':
    seq = "GEEKSFORGEEKS"
    n = len(seq)
    print("The length of the LPS is",
                  lps(seq, 0, n - 1))
     
# This code contributed by Rajput-Ji

C#

// C# program of the above approach
using System;
                     
public class GFG{
 
    // A utility function to get max of two integers
    static int max(int x, int y) {
        return (x > y) ? x : y;
    }
    // Returns the length of the longest palindromic subsequence in seq
  
    static int lps(char []seq, int i, int j) {
    // Base Case 1: If there is only 1 character
        if (i == j) {
            return 1;
        }
  
    // Base Case 2: If there are only 2 characters and both are same
        if (seq[i] == seq[j] && i + 1 == j) {
            return 2;
        }
  
    // If the first and last characters match
        if (seq[i] == seq[j]) {
            return lps(seq, i + 1, j - 1) + 2;
        }
  
    // If the first and last characters do not match
        return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
    }
  
  
    /* Driver program to test above function */
    public static void Main() {
        String seq = "GEEKSFORGEEKS";
        int n = seq.Length;
        Console.Write("The length of the LPS is "+lps(seq.ToCharArray(), 0, n - 1));
  
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

     
    // A utility function to get max of two integers 
    function max(x, y)
    {
        return (x > y) ? x : y;
    }
     
    // Returns the length of the longest palindromic subsequence in seq    
    function lps(seq, i, j)
    {
        // Base Case 1: If there is only 1 character
        if (i == j)
        {
            return 1;
        }
   
        // Base Case 2: If there are only 2 characters and both are same 
            if (seq[i] == seq[j] && i + 1 == j)
            {
                return 2;
            }
       
        // If the first and last characters match 
            if (seq[i] == seq[j])
            {
                return lps(seq, i + 1, j - 1) + 2;
            }
       
        // If the first and last characters do not match 
            return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
    }
     
    /* Driver program to test above function */
    let seq = "GEEKSFORGEEKS";
    let n = seq.length;
    document.write("The length of the LPS is ", lps(seq.split(""), 0, n - 1));
     
    // This code is contributed by avanitrachhadiya2155

---

---

输出：

The length of the LPS is 5

考虑到以上实现，以下是具有所有不同字符的长度为6的序列的部分递归树。

L(0, 5)
             /        \ 
            /          \  
        L(1,5)          L(0,4)
       /    \            /    \
      /      \          /      \
  L(2,5)    L(1,4)  L(1,4)  L(0,3)

在上面的部分递归树中，L(1，4)被求解两次。如果绘制完整的递归树，则可以看到有很多子问题可以一次又一次地解决。由于再次调用了相同的子问题，因此此问题具有“重叠子问题”属性。因此，LPS问题具有动态编程问题的两个属性(请参阅此内容)。像其他典型的动态编程(DP)问题一样，可以通过自下而上的方式构造临时数组L [] []来避免相同子问题的重新计算。
动态编程解决方案

C++

// A Dynamic Programming based C++ program for LPS problem
// Returns the length of the longest palindromic subsequence in seq
#include
#include
 
// A utility function to get max of two integers
int max (int x, int y) { return (x > y)? x : y; }
 
// Returns the length of the longest palindromic subsequence in seq
int lps(char *str)
{
   int n = strlen(str);
   int i, j, cl;
   int L[n][n];  // Create a table to store results of subproblems
 
 
   // Strings of length 1 are palindrome of lentgh 1
   for (i = 0; i < n; i++)
      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/matrix-chain-multiplication-dp-8/). cl is length of
    // substring
    for (cl=2; cl<=n; cl++)
    {
        for (i=0; i

Java

// A Dynamic Programming based Java
// Program for the Egg Dropping Puzzle
class LPS
{
 
    // A utility function to get max of two integers
    static int max (int x, int y) { return (x > y)? x : y; }
     
    // Returns the length of the longest
    // palindromic subsequence in seq
    static int lps(String seq)
    {
    int n = seq.length();
    int i, j, cl;
    // Create a table to store results of subproblems
    int L[][] = new int[n][n];
     
    // Strings of length 1 are palindrome of lentgh 1
    for (i = 0; i < n; i++)
        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/matrix-chain-multiplication-dp-8/).
        // cl is length of substring
        for (cl=2; cl<=n; cl++)
        {
            for (i=0; i

Python

# 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

C＃

// A Dynamic Programming based C# Program
// for the Egg Dropping Puzzle
using System;
 
class GFG {
 
    // A utility function to get max of
    // two integers
    static int max (int x, int y)
    {
        return (x > y)? x : y;
    }
     
    // Returns the length of the longest
    // palindromic subsequence in seq
    static int lps(string seq)
    {
    int n = seq.Length;
    int i, j, cl;
     
    // Create a table to store results
    // of subproblems
    int [,]L = new int[n,n];
     
    // Strings of length 1 are
    // palindrome of lentgh 1
    for (i = 0; i < n; i++)
        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/matrix-chain-multiplication-dp-8/
        // cl is length of substring
        for (cl = 2; cl <= n; cl++)
        {
            for (i = 0; i < n-cl+1; i++)
            {
                j = i + cl - 1;
                 
                if (seq[i] == seq[j] &&
                                  cl == 2)
                    L[i,j] = 2;
                else if (seq[i] == seq[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 */
    public static void Main()
    {
        string seq = "GEEKS FOR GEEKS";
        int n = seq.Length;
        Console.Write("The length of the "
                  + "lps is "+ lps(seq));
    }
}
 
// This code is contributed by nitin mittal.

的PHP

 $y)? $x : $y; }
 
// Returns the length of the
// longest palindromic
// subsequence in seq
function lps($str)
{
$n = strlen($str);
$i; $j; $cl;
 
// Create a table to store
// results of subproblems
$L[][] = array(array());
 
 
// Strings of length 1 are
// palindrome of lentgh 1
for ($i = 0; $i < $n; $i++)
    $L[$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/matrix-chain-multiplication-dp-8/).
    // cl is length of substring
    for ($cl = 2; $cl <= $n; $cl++)
    {
        for ($i = 0; $i < $n - $cl + 1; $i++)
        {
            $j = $i + $cl - 1;
            if ($str[$i] == $str[$j] &&
                            $cl == 2)
            $L[$i][$j] = 2;
            else if ($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 Code
$seq = 'GEEKS FOR GEEKS';
$n = strlen($seq);
echo "The length of the " .
      "LPS is ", lps($seq);
 
// This code is contributed
// by shiv_bhakt.
?>

输出：

The length of the LPS is 7

上述实现的时间复杂度为O(n ^ 2)，这比Naive Recursive实现的最坏情况下的时间复杂度要好得多。