给定由N个长度为M的不同字符串组成的字符串S []数组。任务是通过连接给定数组中的一些字符串来生成最长的回文字符串。
例子：
Input: N = 4, M = 3, S[] = {“omg”, “bbb”, “ffd”, “gmo”}
Output: omgbbbgmo
Explanation: Strings “omg” and “gmo” are reverse of each other and “bbb” is itself a palindrome. Therefore, concatenating “omg” + “bbb” + “gmo” generates the longest palindromic string “omgbbbgmo”.
Input: N = 4, M = 3, s[]={“poy”, “fgh”, “hgf”, “yop”}
Output: poyfghhgfyop
方法：请按照以下步骤解决问题：
- 初始化Set并将每个给定数组中的字符串插入Set中。
- 初始化两个向量left_ans和right_ans，以跟踪获得的回文字符串。
- 现在，遍历字符串数组，并检查其反向是否存在于Set中。
- 如果发现是真的，插入字符串入left_ans之一，另一个为right_ans和擦除无论是从设置，以避免重复的字符串。
- 如果字符串是回文，并且Set中不存在其对，则需要将该字符串附加到结果字符串的中间。
- 打印结果字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
void max_len(string s[], int N, int M)
{
// Stores the distinct strings
// from the given array
unordered_set set_str;
// Insert the strings into set
for (int i = 0; i < N; i++) {
set_str.insert(s[i]);
}
// Stores the left and right
// substrings of the given string
vector left_ans, right_ans;
// Stores the middle substring
string mid;
// Traverse the array of strings
for (int i = 0; i < N; i++) {
string t = s[i];
// Reverse the current string
reverse(t.begin(), t.end());
// Checking if the string is
// itself a palindrome or not
if (t == s[i]) {
mid = t;
}
// Check if the reverse of the
// string is present or not
else if (set_str.find(t)
!= set_str.end()) {
// Append to the left substring
left_ans.push_back(s[i]);
// Append to the right substring
right_ans.push_back(t);
// Erase both the strings
// from the set
set_str.erase(s[i]);
set_str.erase(t);
}
}
// Print the left substring
for (auto x : left_ans) {
cout << x;
}
// Print the middle substring
cout << mid;
reverse(right_ans.begin(),
right_ans.end());
// Print the right substring
for (auto x : right_ans) {
cout << x;
}
}
// Driver Code
int main()
{
int N = 4, M = 3;
string s[] = { "omg", "bbb",
"ffd", "gmo" };
// Function Call
max_len(s, N, M);
return 0;
} Java
// Java program for the
// above approach
import java.util.*;
class GFG{
static String reverse(String input)
{
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--)
{
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
static void max_len(String s[],
int N, int M)
{
// Stores the distinct Strings
// from the given array
HashSet set_str =
new HashSet<>();
// Insert the Strings
// into set
for (int i = 0; i < N; i++)
{
set_str.add(s[i]);
}
// Stores the left and right
// subStrings of the given String
Vector left_ans =
new Vector<>();
Vector right_ans =
new Vector<>();
// Stores the middle
// subString
String mid = "";
// Traverse the array
// of Strings
for (int i = 0; i < N; i++)
{
String t = s[i];
// Reverse the current
// String
t = reverse(t);
// Checking if the String is
// itself a palindrome or not
if (t == s[i])
{
mid = t;
}
// Check if the reverse of the
// String is present or not
else if (set_str.contains(t))
{
// Append to the left
// subString
left_ans.add(s[i]);
// Append to the right
// subString
right_ans.add(t);
// Erase both the Strings
// from the set
set_str.remove(s[i]);
set_str.remove(t);
}
}
// Print the left subString
for (String x : left_ans)
{
System.out.print(x);
}
// Print the middle
// subString
System.out.print(mid);
Collections.reverse(right_ans);
// Print the right subString
for (String x : right_ans)
{
System.out.print(x);
}
}
// Driver Code
public static void main(String[] args)
{
int N = 4, M = 3;
String s[] = {"omg", "bbb",
"ffd", "gmo"};
// Function Call
max_len(s, N, M);
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 program for the above approach
def max_len(s, N, M):
# Stores the distinct strings
# from the given array
set_str = {}
# Insert the strings into set
for i in s:
set_str[i] = 1
# Stores the left and right
# substrings of the given string
left_ans, right_ans = [], []
# Stores the middle substring
mid = ""
# Traverse the array of strings
for i in range(N):
t = s[i]
# Reverse the current string
t = t[::-1]
# Checking if the is
# itself a palindrome or not
if (t == s[i]):
mid = t
# Check if the reverse of the
# is present or not
elif (t in set_str):
# Append to the left substring
left_ans.append(s[i])
# Append to the right substring
right_ans.append(t)
# Erase both the strings
# from the set
del set_str[s[i]]
del set_str[t]
# Print the left substring
for x in left_ans:
print(x, end = "")
# Print the middle substring
print(mid, end = "")
right_ans = right_ans[::-1]
# Print the right substring
for x in right_ans:
print(x, end = "")
# Driver Code
if __name__ == '__main__':
N = 4
M = 3
s = [ "omg", "bbb", "ffd", "gmo"]
# Function call
max_len(s, N, M)
# This code is contributed by mohit kumar 29C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
static String reverse(String input)
{
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--)
{
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("", a);
}
static void max_len(String []s,
int N, int M)
{
// Stores the distinct Strings
// from the given array
HashSet set_str =
new HashSet();
// Insert the Strings
// into set
for (int i = 0; i < N; i++)
{
set_str.Add(s[i]);
}
// Stores the left and right
// subStrings of the given String
List left_ans =
new List();
List right_ans =
new List();
// Stores the middle
// subString
String mid = "";
// Traverse the array
// of Strings
for (int i = 0; i < N; i++)
{
String t = s[i];
// Reverse the current
// String
t = reverse(t);
// Checking if the String is
// itself a palindrome or not
if (t == s[i])
{
mid = t;
}
// Check if the reverse of the
// String is present or not
else if (set_str.Contains(t))
{
// Append to the left
// subString
left_ans.Add(s[i]);
// Append to the right
// subString
right_ans.Add(t);
// Erase both the Strings
// from the set
set_str.Remove(s[i]);
set_str.Remove(t);
}
}
// Print the left subString
foreach (String x in left_ans)
{
Console.Write(x);
}
// Print the middle
// subString
Console.Write(mid);
right_ans.Reverse();
// Print the right subString
foreach (String x in right_ans)
{
Console.Write(x);
}
}
// Driver Code
public static void Main(String[] args)
{
int N = 4, M = 3;
String []s = {"omg", "bbb",
"ffd", "gmo"};
// Function Call
max_len(s, N, M);
}
}
// This code is contributed by 29AjayKumar 输出：
omgbbbgmo
时间复杂度： O(N * M)
辅助空间： O(N * M)