在给定的字符串数组中查找所有回文字符串

给定一个大小为N的字符串数组arr[] ，其中每个字符串仅由小写英文字母组成。任务是查找数组中的所有回文字符串。如果给定数组中不存在回文，则打印 -1。

例子：

Input: arr[] = {“abc”, “car”, “ada”, “racecar”, “cool”}
Output: “ada”, “racecar”
Explanation: These two are the only palindrome strings in the given array

Input: arr[] = {“def”, “ab”}
Output: -1
Explanation: No palindrome string is present in the given array.

方法：解决方案基于贪婪方法。检查数组的每个字符串是否为回文，并跟踪所有回文字符串。请按照以下步骤解决问题：

• 初始化字符串ans的向量。
• 使用变量i遍历[0, N)范围，如果arr[i]是回文，则将其添加到ans中。
• 执行上述步骤后，将存在于ans中的字符串打印为结果字符串。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if given string
// is Palindrome or not
bool isPalindrome(string& s)
{
    // Copy string s char into string a
    string a = s;
    reverse(s.begin(), s.end());
 
    // Check if two string are equal or not
    return s == a;
}
 
// Function to return all Palindrome string
vector PalindromicStrings(string arr[],
                                  int N)
{
    vector ans;
 
    // Loop to find palindrome string
    for (int i = 0; i < N; i++) {
 
        // Checking if given string is
        // palindrome or not
        if (isPalindrome(arr[i])) {
 
            // Update answer variable
            ans.push_back(arr[i]);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
 
    string arr[]
        = { "abc", "car", "ada", "racecar", "cool" };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Print required answer
    vector s = PalindromicStrings(arr, N);
    if(s.size() == 0)
        cout << "-1";
    for(string st: s)
        cout << st << " ";
 
    return 0;
}

Java

// Java code to find the maximum median
// of a sub array having length at least K.
import java.util.*;
public class GFG
{
     
  // Function to check if given string
  // is Palindrome or not
  static boolean isPalindrome(String str)
  {
     
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.length() - 1;
 
    // Keep comparing characters while they are same
    while (h > l)
    {
      if (str.charAt(l++) != str.charAt(h--))
      {
        return false;
      }
    }
    return true;
  }
 
  // Function to return all Palindrome string
  static ArrayList PalindromicStrings(String []arr,
                                      int N)
  {
    ArrayList ans = new ArrayList();
 
    // Loop to find palindrome string
    for (int i = 0; i < N; i++) {
 
      // Checking if given string is
      // palindrome or not
      if (isPalindrome(arr[i])) {
 
        // Update answer variable
        ans.add(arr[i]);
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void main(String args[])
  {
 
    String []arr
      = { "abc", "car", "ada", "racecar", "cool" };
    int N = arr.length;
 
    // Print required answer
    ArrayList s = PalindromicStrings(arr, N);
    if(s.size() == 0)
      System.out.print("-1");
    for (String st : s)
      System.out.print(st + " ");
 
  }
}
 
// This code is contributed by Samim Hossain Mondal.

C#

// C# program for the above approach
using System;
using System.Collections;
class GFG
{
   
  // Function to check if given string
  // is Palindrome or not
  static bool isPalindrome(string str)
  {
     
    // Start from leftmost and rightmost corners of str
    int l = 0;
    int h = str.Length - 1;
 
    // Keep comparing characters while they are same
    while (h > l)
    {
      if (str[l++] != str[h--])
      {
        return false;
      }
    }
    return true;
  }
 
  // Function to return all Palindrome string
  static ArrayList PalindromicStrings(string []arr,
                                      int N)
  {
    ArrayList ans = new ArrayList();
 
    // Loop to find palindrome string
    for (int i = 0; i < N; i++) {
 
      // Checking if given string is
      // palindrome or not
      if (isPalindrome(arr[i])) {
 
        // Update answer variable
        ans.Add(arr[i]);
      }
    }
    return ans;
  }
 
  // Driver Code
  public static void Main()
  {
 
    string []arr
      = { "abc", "car", "ada", "racecar", "cool" };
    int N = arr.Length;
 
    // Print required answer
    ArrayList s = PalindromicStrings(arr, N);
    if(s.Count == 0)
      Console.Write("-1");
    foreach(string st in s)
      Console.Write(st + " ");
 
  }
}
// This code is contributed by Samim Hossain Mondal.

Javascript

 
// JavaScript code for the above approach
 
 
// Function to check if given string
// is Palindrome or not
function isPalindrome(s)
{
    // Copy string s char into string a
    let a = s;
    s = s.split('').reverse().join('');
 
    // Check if two string are equal or not
    return s == a;
}
 
// Function to return all Palindrome string
function PalindromicStrings(arr,N)
{
    let ans = [];
 
    // Loop to find palindrome string
    for (let i = 0; i < N; i++) {
 
        // Checking if given string is
        // palindrome or not
        if (isPalindrome(arr[i])) {
 
            // Update answer variable
            ans.push(arr[i]);
        }
    }
    return ans;
}
 
// Driver Code
let arr = [ "abc", "car", "ada", "racecar", "cool" ];
let N = arr.length;
 
// Print required answer
let s = PalindromicStrings(arr, N);
if(s.length == 0)
    document.write("-1");
for(let st of s)
    document.write(st," ");
 
// This code is contributed by shinjanpatra

输出

ada racecar 

时间复杂度： O(N * W) 其中 W 是字符串的平均长度
辅助空间： O(N * W)