给定一个由N 个长度为M 的不同字符串组成的字符串数组S[] 。任务是通过连接给定数组中的一些字符串来生成可能最长的回文字符串。
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
处理方法：按照以下步骤解决问题：
- 初始化一个 Set 并将给定数组中的每个字符串插入到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。
- 如果一个字符串是一个回文并且它的一对在 Set 中不存在，那么该字符串需要被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)
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