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📜  将字符串拆分为三个回文子串,并尽可能早地进行切割

📅  最后修改于: 2021-09-03 13:51:16             🧑  作者: Mango

给定字符串str ,任务是检查是否可以将给定字符串S拆分为三个回文子串。如果有多个答案是可能的,则打印其中切割索引最少的那个。如果不存在这种可能的分区,则打印“-1”

例子:

处理方法:按照以下步骤解决问题:

  1. 生成字符串的所有可能的子字符串并检查给定的子串是否是回文的。
  2. 如果存在任何子字符串,则将子字符串的最后一个索引存储在向量startPal 中,其中startPal将存储从 0 索引开始并以存储值结束的第一个回文。
  3. 与第一步类似,从给定字符串str的最后一个索引生成所有子字符串,并检查给定子字符串是否为回文。如果任何子字符串作为子字符串存在,则将子字符串的最后一个索引存储在向量lastPal 中,其中lastPal将存储从存储的值开始并在给定字符串str 的(N – 1)索引处结束的第三个回文。
  4. 反转向量lastPal以获得最早的切割。
  5. 现在,迭代两个嵌套循环,外环拾取结束从startPal和内循环索引第一回文主从lastPal第三回文起始索引。如果startPal的值大于lastPal值较小然后将其存储在双形式middlePal矢量。
  6. 现在遍历向量middlePal并检查第一个回文的结束索引和第三个回文的起始索引之间的子串是否为回文。如果发现为真,则第一个回文 = s.substr(0, middlePal[i].first) ,第三个回文 = s.substr(middlePal[i].second, N – middlePal[i].second)和其余字符串将是第三个字符串。
  7. 如果所有三个回文字符串都存在,则打印该字符串,否则打印“-1”

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 a string
// is palindrome or not
bool isPalindrome(string x)
{
    // Copy the string x to y
    string y = x;
 
    // Reverse the string y
    reverse(y.begin(), y.end());
 
    if (x == y) {
 
        // If string x equals y
        return true;
    }
 
    return false;
}
 
// Function to find three palindromes
// from the given string with earliest
// possible cuts
void Palindromes(string S, int N)
{
    // Stores index of palindrome
    // starting from left & right
    vector startPal, lastPal;
 
    string start;
 
    // Store the indices for possible
    // palindromes from left
    for (int i = 0;
         i < S.length() - 2; i++) {
 
        // Push the current character
        start.push_back(S[i]);
 
        // Check for palindrome
        if (isPalindrome(start)) {
 
            // Insert the current index
            startPal.push_back(i);
        }
    }
 
    string last;
 
    // Stores the indexes for possible
    // palindromes from right
    for (int j = S.length() - 1;
         j >= 2; j--) {
 
        // Push the current character
        last.push_back(S[j]);
 
        // Check palindromic
        if (isPalindrome(last)) {
 
            // Insert the current index
            lastPal.push_back(j);
        }
    }
 
    // Sort the indexes for palindromes
    // from right in ascending order
    reverse(lastPal.begin(),
            lastPal.end());
 
    vector > middlePal;
 
    for (int i = 0;
         i < startPal.size(); i++) {
 
        for (int j = 0;
             j < lastPal.size(); j++) {
 
            // If the value of startPal
            // < lastPal value then
            // store it in middlePal
            if (startPal[i] < lastPal[j]) {
 
                // Insert the current pair
                middlePal.push_back(
                    { startPal[i],
                      lastPal[j] });
            }
        }
    }
 
    string res1, res2, res3;
    int flag = 0;
 
    // Traverse over the middlePal
    for (int i = 0;
         i < middlePal.size(); i++) {
 
        int x = middlePal[i].first;
        int y = middlePal[i].second;
 
        string middle;
 
        for (int k = x + 1; k < y; k++) {
            middle.push_back(S[k]);
        }
 
        // Check if the middle part
        // is palindrome
        if (isPalindrome(middle)) {
            flag = 1;
            res1 = S.substr(0, x + 1);
            res2 = middle;
            res3 = S.substr(y, N - y);
            break;
        }
    }
 
    // Print the three palindromic
    if (flag == 1) {
        cout << res1 << " "
             << res2 << " "
             << res3;
    }
 
    // Otherwise
    else
        cout << "-1";
}
 
// Driver Code
int main()
{
    // Given string str
    string str = "ababbcb";
 
    int N = str.length();
 
    // Function Call
    Palindromes(str, N);
 
    return 0;
}


Java
// Java program for the
// above approach
import java.util.*;
class GFG{
 
static class pair
{
  int first, second;
  public pair(int first,
              int second)
  {
    this.first = first;
    this.second = second;
  }
}
 
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);
}
 
// Function to check if a String
// is palindrome or not
static boolean isPalindrome(String x)
{
  // Copy the String x to y
  String y = x;
 
  // Reverse the String y
  y = reverse(y);
 
  if (x.equals(y))
  {
    // If String x equals y
    return true;
  }
 
  return false;
}
 
// Function to find three palindromes
// from the given String with earliest
// possible cuts
static void Palindromes(String S,
                        int N)
{
  // Stores index of palindrome
  // starting from left & right
  Vector startPal =
         new Vector<>();
  Vector lastPal =
         new Vector<>();
 
  String start = "";
 
  // Store the indices for possible
  // palindromes from left
  for (int i = 0;
           i < S.length() - 2; i++)
  {
    // Push the current character
    start += S.charAt(i);
 
    // Check for palindrome
    if (isPalindrome(start))
    {
      // Insert the current index
      startPal.add(i);
    }
  }
 
  String last = "";
 
  // Stores the indexes for possible
  // palindromes from right
  for (int j = S.length() - 1;
           j >= 2; j--)
  {
    // Push the current character
    last += S.charAt(j);
 
    // Check palindromic
    if (isPalindrome(last))
    {
      // Insert the current index
      lastPal.add(j);
    }
  }
 
  // Sort the indexes for palindromes
  // from right in ascending order
  Collections.reverse(lastPal);
 
  Vector middlePal =
               new Vector<>();
   
  for (int i = 0;
           i < startPal.size(); i++)
  {
    for (int j = 0;
             j < lastPal.size(); j++)
    {
      // If the value of startPal
      // < lastPal value then
      // store it in middlePal
      if (startPal.get(i) <
          lastPal.get(j))
      {
        // Insert the current pair
        middlePal.add(new pair(startPal.get(i),
                               lastPal.get(j)));
      }
    }
  }
 
  String res1 = "",
         res2 = "", res3 = "";
  int flag = 0;
 
  // Traverse over the middlePal
  for (int i = 0;
           i < middlePal.size(); i++)
  {
    int x = middlePal.get(i).first;
    int y = middlePal.get(i).second;
    String middle = "";
 
    for (int k = x + 1; k < y; k++)
    {
      middle += S.charAt(k);
    }
 
    // Check if the middle part
    // is palindrome
    if (isPalindrome(middle))
    {
      flag = 1;
      res1 = S.substring(0, x + 1);
      res2 = middle;
      res3 = S.substring(y, N);
      break;
    }
  }
 
  // Print the three palindromic
  if (flag == 1)
  {
    System.out.print(res1 + " " +
                     res2 + " " + res3);
  }
 
  // Otherwise
  else
    System.out.print("-1");
}
 
// Driver Code
public static void main(String[] args)
{
  // Given String str
  String str = "ababbcb";
 
  int N = str.length();
 
  // Function Call
  Palindromes(str, N);
}
}
 
// This code is contributed by shikhasingrajput


Python3
# Python3 program for the above approach
 
# Function to check if a strring
# is palindrome or not
def isPalindrome(x):
     
    # Copy the x to y
    y = x
 
    # Reverse the y
    y = y[::-1]
 
    if (x == y):
 
        # If x equals y
        return True
         
    return False
 
# Function to find three palindromes
# from the given with earliest
# possible cuts
def Palindromes(S, N):
     
    # Stores index of palindrome
    # starting from left & right
    startPal, lastPal = [], []
 
    start = []
 
    # Store the indices for possible
    # palindromes from left
    for i in range(len(S) - 2):
 
        # Push the current character
        start.append(S[i])
 
        # Check for palindrome
        if (isPalindrome(start)):
 
            # Insert the current index
            startPal.append(i)
 
    last = []
 
    # Stores the indexes for possible
    # palindromes from right
    for j in range(len(S) - 1, 1, -1):
 
        # Push the current character
        last.append(S[j])
 
        # Check palindromic
        if (isPalindrome(last)):
 
            # Insert the current index
            lastPal.append(j)
 
    # Sort the indexes for palindromes
    # from right in ascending order
    lastPal = lastPal[::-1]
 
    middlePal = []
 
    for i in range(len(startPal)):
        for j in range(len(lastPal)):
 
            # If the value of startPal
            # < lastPal value then
            # store it in middlePal
            if (startPal[i] < lastPal[j]):
 
                # Insert the current pair
                middlePal.append([startPal[i],
                                   lastPal[j]])
 
    res1, res2, res3 = "", "", ""
    flag = 0
 
    # Traverse over the middlePal
    for i in range(len(middlePal)):
        x = middlePal[i][0]
        y = middlePal[i][1]
        #print(x,y)
 
        middle = ""
 
        for k in range(x + 1, y):
            middle += (S[k])
 
        # Check if the middle part
        # is palindrome
        if (isPalindrome(middle)):
            flag = 1
            res1 = S[0 : x + 1]
            res2 = middle
            res3 = S[y : N]
            #print(S,x,y,N)
            break
 
    # Print the three palindromic
    if (flag == 1):
        print(res1, res2, res3)
 
    # Otherwise
    else:
        print("-1")
 
# Driver Code
if __name__ == '__main__':
     
    # Given strr
    strr = "ababbcb"
 
    N = len(strr)
 
    # Function call
    Palindromes(strr, N)
 
# This code is contributed by mohit kumar 29


C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
public class pair
{
  public int first, second;
  public pair(int first,
              int second)
  {
    this.first = first;
    this.second = second;
  }
}
 
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);
}
 
// Function to check if a String
// is palindrome or not
static bool isPalindrome(String x)
{
   
  // Copy the String x to y
  String y = x;
 
  // Reverse the String y
  y = reverse(y);
 
  if (x.Equals(y))
  {
     
    // If String x equals y
    return true;
  }
  return false;
}
 
// Function to find three palindromes
// from the given String with earliest
// possible cuts
static void Palindromes(String S,
                        int N)
{
   
  // Stores index of palindrome
  // starting from left & right
  List startPal = new List();
  List lastPal = new List();
   
  String start = "";
 
  // Store the indices for possible
  // palindromes from left
  for(int i = 0;
          i < S.Length - 2; i++)
  {
     
    // Push the current character
    start += S[i];
 
    // Check for palindrome
    if (isPalindrome(start))
    {
       
      // Insert the current index
      startPal.Add(i);
    }
  }
 
  String last = "";
 
  // Stores the indexes for possible
  // palindromes from right
  for(int j = S.Length - 1;
          j >= 2; j--)
  {
     
    // Push the current character
    last += S[j];
 
    // Check palindromic
    if (isPalindrome(last))
    {
       
      // Insert the current index
      lastPal.Add(j);
    }
  }
   
  // Sort the indexes for palindromes
  // from right in ascending order
  lastPal.Reverse();
 
  List middlePal = new List();
   
  for(int i = 0;
          i < startPal.Count; i++)
  {
    for(int j = 0;
            j < lastPal.Count; j++)
    {
       
      // If the value of startPal
      // < lastPal value then
      // store it in middlePal
      if (startPal[i] < lastPal[j])
      {
         
        // Insert the current pair
        middlePal.Add(new pair(startPal[i],
                                lastPal[j]));
      }
    }
  }
 
  String res1 = "", res2 = "",
         res3 = "";
  int flag = 0;
 
  // Traverse over the middlePal
  for(int i = 0;
          i < middlePal.Count; i++)
  {
    int x = middlePal[i].first;
    int y = middlePal[i].second;
    String middle = "";
 
    for(int k = x + 1; k < y; k++)
    {
      middle += S[k];
    }
 
    // Check if the middle part
    // is palindrome
    if (isPalindrome(middle))
    {
      flag = 1;
      res1 = S.Substring(0, x + 1);
      res2 = middle;
      res3 = S.Substring(y);
      break;
    }
  }
 
  // Print the three palindromic
  if (flag == 1)
  {
    Console.Write(res1 + " " +
                  res2 + " " + res3);
  }
 
  // Otherwise
  else
    Console.Write("-1");
}
 
// Driver Code
public static void Main(String[] args)
{
   
  // Given String str
  String str = "ababbcb";
 
  int N = str.Length;
 
  // Function Call
  Palindromes(str, N);
}
}
 
// This code is contributed by Amit Katiyar


输出:
a bab bcb






时间复杂度: O(N 2 )
辅助空间: O(N)

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