将一个字符串分割成至少长度为 2 的回文字符串，每个字符出现在一个字符串

给定一个由N个小写字母组成的字符串S ，任务是检查所有长度至少为 2的字符串是否通过只选择字符串S 的每个字符一次而形成的所有字符串都是回文的。如果发现是真的，则打印“是” 。否则，打印“否” 。

例子：

Input: S = “abbbaddzcz”
Output: Yes
Explanation: The palindromic strings of length greater than 1 are { “aa”, “bbb”, “dd”, “zcz”} and all the characters in the given string is used only once. Therefore, print “Yes”.

Input: S = “abcd”
Output: No

方法：思路是将字符串分解成偶数长度的回文字符串，如果存在频率为1的字符，则将其与偶数长度的回文字符串配对。请按照以下步骤解决问题：

• 初始化一个数组，比如freq[] ，大小为26 ，以存储字符串中每个字符的频率。
• 迭代给定字符串S的字符并更新数组freq[]中每个字符的频率。
• 初始化两个变量，说O和E，存储分别为偶数频率独特的和计数。
• 遍历数组freq[]并且如果freq[i]等于1 ，则将O增加1 。否则，将E增加1 。
• 检查是否E ≥ O ，然后打印“是” 。否则，请执行以下步骤：
  • Ø更新至O – E，以配对后的剩余独特的字符存储。
  • 遍历数组freq[] ，如果O的值最多为0 ，则跳出循环。否则，更新O操作-O – (FREQ [I] / 2)，并通过1然后增量O操作。
• 完成上述步骤后，如果O ≤ 0 ，则打印“是” 。否则，打印“否” 。

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 can be
// split into palindromic strings of at
// least length 2 by including every
// character exactly once
void checkPalindrome(string& s)
{
    // Stores the frequency
    // of each character
    int a[26] = { 0 };
 
    // Store the frequency of
    // characters with frequencies
    // 1 and even respectively
    int o = 0, e = 0;
 
    // Traverse the string s
    for (int i = 0; s[i] != '\0'; i++)
        a[(int)s[i] - 97]++;
 
    // Iterate over all the characters
    for (int i = 0; i < 26; i++) {
 
        // If the frequency is 1
        if (a[i] == 1)
            o++;
 
        // If frequency is even
        else if (a[i] % 2 == 0
                 and a[i] != 0)
            e += (a[i] / 2);
    }
 
    // Print the result
    if (e >= o)
        cout << "Yes";
 
    else {
 
        // Stores the number of characters
        // with frequency equal to 1 that
        // are not part of a palindromic string
        o = o - e;
 
        // Iterate over all the characters
        for (int i = 0; i < 26; i++) {
 
            // If o becomes less than 0,
            // then break out of the loop
            if (o <= 0)
                break;
 
            // If frequency of the current
            // character is > 2 and is odd
            if (o > 0
                and a[i] % 2 == 1
                and a[i] > 2) {
 
                int k = o;
 
                // Update the value of o
                o = o - a[i] / 2;
 
                // If a single character
                // is still remaining
                if (o > 0 or 2 * k + 1 == a[i]) {
 
                    // Increment o by 1
                    o++;
 
                    // Set a[i] to 1
                    a[i] = 1;
                }
            }
        }
 
        // Print the result
        if (o <= 0)
            cout << "Yes";
        else
            cout << "No";
    }
}
 
// Driver Code
int main()
{
    string S = "abbbaddzcz";
    checkPalindrome(S);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
// Function to check if  a string can be
// split into palindromic strings of at
// least length 2 by including every
// character exactly once
static void checkPalindrome(String s)
{
     
    // Stores the frequency
    // of each character
    int a[] = new int[26];
 
    // Store the frequency of
    // characters with frequencies
    // 1 and even respectively
    int o = 0, e = 0;
 
    // Traverse the string s
    for(int i = 0; i < s.length(); i++)
        a[s.charAt(i) - 'a']++;
 
    // Iterate over all the characters
    for(int i = 0; i < 26; i++)
    {
         
        // If the frequency is 1
        if (a[i] == 1)
            o++;
 
        // If frequency is even
        else if (a[i] % 2 == 0 && a[i] != 0)
            e += (a[i] / 2);
    }
 
    // Print the result
    if (e >= o)
        System.out.println("Yes");
 
    else
    {
         
        // Stores the number of characters
        // with frequency equal to 1 that
        // are not part of a palindromic string
        o = o - e;
 
        // Iterate over all the characters
        for(int i = 0; i < 26; i++)
        {
             
            // If o becomes less than 0,
            // then break out of the loop
            if (o <= 0)
                break;
 
            // If frequency of the current
            // character is > 2 and is odd
            if (o > 0 && a[i] % 2 == 1 && a[i] > 2)
            {
                int k = o;
 
                // Update the value of o
                o = o - a[i] / 2;
 
                // If a single character
                // is still remaining
                if (o > 0 || 2 * k + 1 == a[i])
                {
                     
                    // Increment o by 1
                    o++;
 
                    // Set a[i] to 1
                    a[i] = 1;
                }
            }
        }
 
        // Print the result
        if (o <= 0)
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    String S = "abbbaddzcz";
     
    checkPalindrome(S);
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
 
# Function to check if  a string can be
# split into palindromic strings of at
# least length 2 by including every
# character exactly once
def checkPalindrome(s):
     
    # Stores the frequency
    # of each character
    a = [0] * 26
 
    # Store the frequency of
    # characters with frequencies
    # 1 and even respectively
    o, e = 0, 0
 
    # Traverse the string s
    for i in s:
        a[ord(i) - 97] += 1
 
    # Iterate over all the characters
    for i in range(26):
         
        # If the frequency is 1
        if (a[i] == 1):
            o += 1
 
        # If frequency is even
        elif (a[i] % 2 == 0 and a[i] != 0):
            e += (a[i] // 2)
 
    # Print the result
    if (e >= o):
        print("Yes")
    else:
         
        # Stores the number of characters
        # with frequency equal to 1 that
        # are not part of a palindromic string
        o = o - e
 
        # Iterate over all the characters
        for i in range(26):
             
            # If o becomes less than 0,
            # then break out of the loop
            if (o <= 0):
                break
 
            # If frequency of the current
            # character is > 2 and is odd
            if (o > 0 and a[i] % 2 == 1 and a[i] > 2):
                k = o
                 
                # Update the value of o
                o = o - a[i] // 2
 
                # If a single character
                # is still remaining
                if (o > 0 or 2 * k + 1 == a[i]):
 
                    # Increment o by 1
                    o += 1
 
                    # Set a[i] to 1
                    a[i] = 1
 
        # Print the result
        if (o <= 0):
            print("Yes")
        else:
            print("No")
             
# Driver Code
if __name__ == '__main__':
     
    S = "abbbaddzcz"
     
    checkPalindrome(S)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if  a string can be
// split into palindromic strings of at
// least length 2 by including every
// character exactly once
static void checkPalindrome(string s)
{
     
    // Stores the frequency
    // of each character
    int[] a = new int[26];
 
    // Store the frequency of
    // characters with frequencies
    // 1 and even respectively
    int o = 0, e = 0;
 
    // Traverse the string s
    for(int i = 0; i < s.Length; i++)
        a[s[i] - 'a']++;
 
    // Iterate over all the characters
    for(int i = 0; i < 26; i++)
    {
         
        // If the frequency is 1
        if (a[i] == 1)
            o++;
 
        // If frequency is even
        else if (a[i] % 2 == 0 && a[i] != 0)
            e += (a[i] / 2);
    }
 
    // Print the result
    if (e >= o)
        Console.WriteLine("Yes");
 
    else
    {
         
        // Stores the number of characters
        // with frequency equal to 1 that
        // are not part of a palindromic string
        o = o - e;
 
        // Iterate over all the characters
        for(int i = 0; i < 26; i++)
        {
             
            // If o becomes less than 0,
            // then break out of the loop
            if (o <= 0)
                break;
 
            // If frequency of the current
            // character is > 2 and is odd
            if (o > 0 && a[i] % 2 == 1 && a[i] > 2)
            {
                int k = o;
 
                // Update the value of o
                o = o - a[i] / 2;
 
                // If a single character
                // is still remaining
                if (o > 0 || 2 * k + 1 == a[i])
                {
                     
                    // Increment o by 1
                    o++;
 
                    // Set a[i] to 1
                    a[i] = 1;
                }
            }
        }
 
        // Print the result
        if (o <= 0)
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// Driver code
static void Main()
{
    string S = "abbbaddzcz";
     
    checkPalindrome(S);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
        // Javascript program for the above approach
 
        // Function to check if  a string can be
        // split into palindromic strings of at
        // least length 2 by including every
        // character exactly once
        function checkPalindrome( s )
        {
            // Stores the frequency
            // of each character
            let a = [0]
 
            // Store the frequency of
            // characters with frequencies
            // 1 and even respectively
            let o = 0, e = 0;
 
            // Traverse the string s
            for (let i = 0; i<s.length; i++)
                a[s.charCodeAt(i) - 97]++;
 
            // Iterate over all the characters
            for (let i = 0; i < 26; i++) {
 
                // If the frequency is 1
                if (a[i] == 1)
                    o++;
 
                // If frequency is even
                else if (a[i] % 2 == 0 && a[i] != 0)
                    e += (a[i] / 2);
            }
 
            // Print the result
            if (e >= o)
                document.write("Yes")
 
            else {
 
                // Stores the number of characters
                // with frequency equal to 1 that
                // are not part of a palindromic string
                o = o - e;
 
                // Iterate over all the characters
                for (let i = 0; i < 26; i++)
                {
 
                    // If o becomes less than 0,
                    // then break out of the loop
                    if (o <= 0)
                        break;
 
                    // If frequency of the current
                    // character is > 2 and is odd
                    if (o > 0 && a[i] % 2 == 1 && a[i] > 2)
                    {
 
                        let k = o;
 
                        // Update the value of o
                        o = o - a[i] / 2;
 
                        // If a single character
                        // is still remaining
                        if (o > 0 || 2 * k + 1 == a[i])
                        {
 
                            // Increment o by 1
                            o++;
 
                            // Set a[i] to 1
                            a[i] = 1;
                        }
                    }
                }
 
                // Print the result
                if (o <= 0)
                document.write("Yes")
                else
                document.write("No")
            }
        }
 
        // Driver Code
         
        let S = "abbbaddzcz";
        checkPalindrome(S);
 
        // This code is contributed by Hritik
         
    

输出：

Yes

时间复杂度： O(N)
辅助空间： O(1)

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