给定一个由小写字母组成的字符串S ，任务是找到仅由两个不同字符组成的最长回文子序列的长度。
例子：
Input: S = “bbccdcbb”
Output: 7
Explanation:
The longest palindromic subsequence of the desired form is “bbcccbb”, which is of length 7.
Input: S = “aeea”
Output: 4
Explanation:
The longest palindromic subsequence of desired form is “aeea”, which is of length 4.
方法：
为了解决该问题，我们需要执行以下步骤：
- 将每个字符的出现次数存储在一个前缀数组中，并将每个字符的位置存储在另一个数组中。
- 现在对于每对等于字符的计算它们之间的字符的出现的最大数。
下面是上述方法的实现：
C++
// C++ implementation to find Longest
// Palindromic Subsequence consisting
// of two distinct characters only
#include
using namespace std;
// Function that prints the length of
// maximum required subsequence
void longestPalindrome(string s)
{
// Calculate length of string
int n = s.length();
vector > pref(
26,
vector(n, 0));
vector > pos(26);
pref[s[0] - 'a'][0]++;
pos[s[0] - 'a'].push_back(0);
// Store number of occurrences of each
// character and position of each
// chatacter in string
for (int i = 1; i < n; i++) {
for (int j = 0; j < 26; j++)
pref[j][i] += pref[j][i - 1];
int index = s[i] - 'a';
pref[index][i]++;
pos[index].push_back(i);
}
int ans = 0;
// Iterate all characters
for (int i = 0; i < 26; i++) {
// Calculate number of
// occurences of the
// current character
int size = pos[i].size();
ans = max(ans, size);
// Iterate half of the
// number of positions
// of current character
for (int j = 0; j < size / 2; j++) {
int l = pos[i][j];
int r = pos[i][size - j - 1] - 1;
// Determine maximum length
// of a character between
// l and r position
for (int k = 0; k < 26; k++) {
int sum = pref[k][r] - pref[k][l];
// Compute the maximum from all
ans = max(ans, 2 * (j + 1) + sum);
}
}
}
// Printing maximum length
cout << ans << "\n";
}
// Driver Code
int main()
{
string S = "bbccdcbb";
longestPalindrome(S);
return 0;
} Java
// Java implementation to find Longest
// Palindromic Subsequence consisting
// of two distinct characters only
import java.util.*;
class GFG{
// Function that prints the length of
// maximum required subsequence
static void longestPalindrome(String s)
{
// Calculate length of String
int n = s.length();
int [][]pref = new int[26][n];
Vector []pos = new Vector[26];
for (int i = 0; i < pos.length; i++)
pos[i] = new Vector();
pref[s.charAt(0) - 'a'][0]++;
pos[s.charAt(0) - 'a'].add(0);
// Store number of occurrences of each
// character and position of each
// chatacter in String
for (int i = 1; i < n; i++)
{
for (int j = 0; j < 26; j++)
pref[j][i] += pref[j][i - 1];
int index = s.charAt(i) - 'a';
pref[index][i]++;
pos[index].add(i);
}
int ans = 0;
// Iterate all characters
for (int i = 0; i < 26; i++)
{
// Calculate number of
// occurences of the
// current character
int size = pos[i].size();
ans = Math.max(ans, size);
// Iterate half of the
// number of positions
// of current character
for (int j = 0; j < size / 2; j++)
{
int l = pos[i].elementAt(j);
int r = pos[i].elementAt(size - j - 1) - 1;
// Determine maximum length
// of a character between
// l and r position
for (int k = 0; k < 26; k++)
{
int sum = pref[k][r] - pref[k][l];
// Compute the maximum from all
ans = Math.max(ans, 2 *
(j + 1) + sum);
}
}
}
// Printing maximum length
System.out.print(ans + "\n");
}
// Driver Code
public static void main(String[] args)
{
String S = "bbccdcbb";
longestPalindrome(S);
}
}
// This code is contributed by Princi Singh C#
// C# implementation to find longest
// Palindromic Subsequence consisting
// of two distinct characters only
using System;
using System.Collections.Generic;
class GFG{
// Function that prints
// the length of maximum
// required subsequence
static void longestPalindrome(String s)
{
// Calculate length of String
int n = s.Length;
int [,]pref = new int[26, n];
List []pos = new List[26];
for (int i = 0; i < pos.Length; i++)
pos[i] = new List();
pref[s[0] - 'a', 0]++;
pos[s[0] - 'a'].Add(0);
// Store number of occurrences of each
// character and position of each
// chatacter in String
for (int i = 1; i < n; i++)
{
for (int j = 0; j < 26; j++)
pref[j, i] += pref[j, i - 1];
int index = s[i] - 'a';
pref[index, i]++;
pos[index].Add(i);
}
int ans = 0;
// Iterate all characters
for (int i = 0; i < 26; i++)
{
// Calculate number of
// occurences of the
// current character
int size = pos[i].Count;
ans = Math.Max(ans, size);
// Iterate half of the
// number of positions
// of current character
for (int j = 0; j < size / 2; j++)
{
int l = pos[i][j];
int r = pos[i][size -
j - 1] - 1;
// Determine maximum length
// of a character between
// l and r position
for (int k = 0; k < 26; k++)
{
int sum = pref[k, r] -
pref[k, l];
// Compute the maximum from all
ans = Math.Max(ans, 2 *
(j + 1) + sum);
}
}
}
// Printing maximum length
Console.Write(ans + "\n");
}
// Driver Code
public static void Main(String[] args)
{
String S = "bbccdcbb";
longestPalindrome(S);
}
}
// This code is contributed by Amit Katiyar 输出：
7