长度为 2 或更长的重复子序列
给定一个字符串，查找是否有任何长度为 2 或更多的子序列重复自身，使得两个子序列在同一位置没有相同的字符，即两个子序列中的任何第 0 或第 1 个字符都应该t 在原始字符串中具有相同的索引。
例子：
Input: ABCABD
Output: Repeated Subsequence Exists (A B is repeated)
Input: ABBB
Output: Repeated Subsequence Exists (B B is repeated)
Input: AAB
Output: Repeated Subsequence Doesn't Exist (Note that
A B cannot be considered as repeating because B is at
same position in two subsequences).
Input: AABBC
Output: Repeated Subsequence Exists (A B is repeated)
Input: ABCDACB
Output: Repeated Subsequence Exists (A B is repeated)
Input: ABCD
Output: Repeated Subsequence Doesn't Exist该问题是最长公共子序列问题的经典变体。我们在这里讨论了动态编程解决方案。动态规划解决方案需要 O(n 2 ) 时间和空间。
在这篇文章中，讨论了 O(n) 时间和空间方法。
这个想法是从字符串中删除所有非重复字符并检查结果字符串是否为回文。如果剩下的字符串是回文，则不重复，否则有重复。对于像“AAA”这样的输入，我们需要处理一种特殊情况，这些输入是回文，但存在重复的子序列。如果回文字符串的长度为奇数并且其中间字母与左（或右）字符相同，则它存在重复子序列。
下面是上述想法的实现。
C++
// C++ program to check if any repeated
// subsequence exists in the string
#include
#define MAX_CHAR 256
using namespace std;
// A function to check if a string str is palindrome
bool isPalindrome(char str[], int l, int h)
{
// l and h are leftmost and rightmost corners of str
// Keep comparing characters while they are same
while (h > l)
if (str[l++] != str[h--])
return false;
return true;
}
// The main function that checks if repeated
// subsequence exists in the string
int check(char str[])
{
// Find length of input string
int n = strlen(str);
// Create an array to store all characters and their
// frequencies in str[]
int freq[MAX_CHAR] = { 0 };
// Traverse the input string and store frequencies
// of all characters in freq[] array.
for (int i = 0; i < n; i++)
{
freq[str[i]]++;
// If the character count is more than 2
// we found a repetition
if (freq[str[i]] > 2)
return true;
}
// In-place remove non-repeating characters
// from the string
int k = 0;
for (int i = 0; i < n; i++)
if (freq[str[i]] > 1)
str[k++] = str[i];
str[k] = '\0';
// check if the resultant string is palindrome
if (isPalindrome(str, 0, k-1))
{
// special case - if length is odd
// return true if the middle character is
// same as previous one
if (k & 1)
return str[k/2] == str[k/2 - 1];
// return false if string is a palindrome
return false;
}
// return true if string is not a palindrome
return true;
}
// Driver code
int main()
{
char str[] = "ABCABD";
if (check(str))
cout << "Repeated Subsequence Exists";
else
cout << "Repeated Subsequence Doesn't Exists";
return 0;
} Java
// Java program to check if any repeated
// subsequence exists in the string
class GFG
{
static int MAX_CHAR = 256;
// A function to check
// if a string str is palindrome
public static boolean isPalindrome(String str,
int l, int h)
{
// l and h are leftmost and rightmost corners of str
// Keep comparing characters while they are same
while(h > l)
if (str.charAt(l++) != str.charAt(h--))
return false;
return true;
}
// The main function that checks if repeated
// subsequence exists in the string
public static boolean check(String str)
{
// Find length of input string
int n = str.length();
// Create an array to store all characters
// and their frequencies in str[]
int[] freq = new int[MAX_CHAR];
// Traverse the input string and store frequencies
// of all characters in freq[] array.
for (int i = 0; i < n; i++)
{
freq[str.charAt(i)]++;
// If the character count is more than 2
// we found a repetition
if (freq[str.charAt(i)] > 2)
return true;
}
// In-place remove non-repeating characters
// from the string
int k = 0;
for (int i = 0; i < n; i++)
if (freq[str.charAt(i)] > 1)
str.replace(str.charAt(k++),
str.charAt(i));
str.replace(str.charAt(k), '\0');
// check if the resultant string is palindrome
if (isPalindrome(str, 0, k - 1))
{
// special case - if length is odd
// return true if the middle character is
// same as previous one
if ((k & 1) == 1)
{
// It is checked so that
// StringIndexOutOfBounds can be avoided
if (k / 2 >= 1)
return (str.charAt(k / 2) ==
str.charAt(k / 2 - 1));
}
// return false if string is a palindrome
return false;
}
// return true if string is not a palindrome
return true;
}
// Driver Code
public static void main(String[] args)
{
String str = "ABCABD";
if (check(str))
System.out.println("Repeated Subsequence Exists");
else
System.out.println("Repeated Subsequence" +
" Doesn't Exists");
}
}
// This code is contributed by sanjeev2552Python3
# Python3 program to check if any repeated
# subsequence exists in the String
MAX_CHAR = 256
# A function to check
# if a String Str is palindrome
def isPalindrome(Str, l, h):
# l and h are leftmost and rightmost corners of Str
# Keep comparing characters while they are same
while (h > l):
if (Str[l] != Str[h]):
l += 1
h -= 1
return False
return True
# The main function that checks if repeated
# subsequence exists in the String
def check(Str):
# Find length of input String
n = len(Str)
# Create an array to store all characters
# and their frequencies in Str[]
freq = [0 for i in range(MAX_CHAR)]
# Traverse the input String and
# store frequencies of all characters
# in freq[] array.
for i in range(n):
freq[ord(Str[i])] += 1
# If the character count is more than 2
# we found a repetition
if (freq[ord(Str[i])] > 2):
return True
# In-place remove non-repeating characters
# from the String
k = 0
for i in range(n):
if (freq[ord(Str[i])] > 1):
Str[k] = Str[i]
k += 1
Str[k] = '\0'
# check if the resultant String is palindrome
if (isPalindrome(Str, 0, k - 1)):
# special case - if length is odd
# return true if the middle character is
# same as previous one
if (k & 1):
return Str[k // 2] == Str[k // 2 - 1]
# return false if String is a palindrome
return False
# return true if String is not a palindrome
return True
# Driver code
S = "ABCABD"
Str = [i for i in S]
if (check(Str)):
print("Repeated Subsequence Exists")
else:
print("Repeated Subsequence Doesn't Exists")
# This code is contributed by Mohit KumarC#
// C# program to check if any repeated
// subsequence exists in the string
using System;
class GFG
{
static int MAX_CHAR = 256;
// A function to check
// if a string str is palindrome
public static Boolean isPalindrome(String str,
int l, int h)
{
// l and h are leftmost and rightmost corners of str
// Keep comparing characters while they are same
while(h > l)
if (str[l++] != str[h--])
return false;
return true;
}
// The main function that checks if repeated
// subsequence exists in the string
public static Boolean check(String str)
{
// Find length of input string
int n = str.Length;
// Create an array to store all characters
// and their frequencies in str[]
int[] freq = new int[MAX_CHAR];
// Traverse the input string and store frequencies
// of all characters in freq[] array.
for (int i = 0; i < n; i++)
{
freq[str[i]]++;
// If the character count is more than 2
// we found a repetition
if (freq[str[i]] > 2)
return true;
}
// In-place remove non-repeating characters
// from the string
int k = 0;
for (int i = 0; i < n; i++)
if (freq[str[i]] > 1)
str.Replace(str[k++],
str[i]);
str.Replace(str[k], '\0');
// check if the resultant string is palindrome
if (isPalindrome(str, 0, k - 1))
{
// special case - if length is odd
// return true if the middle character is
// same as previous one
if ((k & 1) == 1)
{
// It is checked so that
// StringIndexOutOfBounds can be avoided
if (k / 2 >= 1)
return (str[k / 2] ==
str[k / 2 - 1]);
}
// return false if string is a palindrome
return false;
}
// return true if string is not a palindrome
return true;
}
// Driver Code
public static void Main(String[] args)
{
String str = "ABCABD";
if (check(str))
Console.WriteLine("Repeated Subsequence Exists");
else
Console.WriteLine("Repeated Subsequence" +
" Doesn't Exists");
}
}
// This code is contributed by Princi SinghJavascript
输出 ：
Repeated Subsequence Exists