检查字符串是否遵循 a^nb^n 模式

给定字符串str，返回 true字符串遵循模式 a ⁿ b ⁿ ，即它有 a 后跟 b ，因此 a 和 b 的数量相同。

例子：

Input : str = "aabb"
Output : Yes

Input : str = "abab"
Output : No

Input : str = "aabbb"
Output : No

这个想法是首先计算a。如果 a 的数量不等于字符串长度的一半，则返回 false。否则检查是否所有剩余的字符都是 b。

下面是上述想法的实现：

C++

// C++ program to check if a string is of
// the form a^nb^n.
#include 
using namespace std;
 
// Returns true str is of the form a^nb^n.
bool isAnBn(string str)
{
    int n = str.length();
 
    // After this loop 'i' has count of a's
    int i;
    for (i = 0; i < n; i++)
        if (str[i] != 'a')
            break;
 
    // Since counts of a's and b's should
    // be equal, a should appear exactly
    // n/2 times
    if (i * 2 != n)
        return false;
 
    // Rest of the characters must be all 'b'
    int j;
    for (j = i; j < n; j++)
        if (str[j] != 'b')
            return false;
 
    return true;
}
 
// Driver code
int main()
{
    string str = "abab";
   
    // Function call
    isAnBn(str) ? cout << "Yes" : cout << "No";
    return 0;
}

Java

// Java program to check if a string is of
// the form a^nb^n.
import java.util.*;
import java.lang.*;
import java.io.*;
 
class CheckPattern {
    public static boolean isAnBn(String s)
    {
        int l = s.length();
 
        // Only even length strings will have same number of
        // a's and b's
        if (l % 2 == 1) {
            return false;
        }
        // Set two pointers, one from the left and another
        // from right
        int i = 0;
        int j = l - 1;
 
        // Compare the characters till the center
        while (i < j) {
            if (s.charAt(i) != 'a' || s.charAt(j) != 'b') {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
 
    public static void main(String[] args)
        throws java.lang.Exception
    {
        String s = "abab";
 
        // Function call
        boolean value = isAnBn(s);
        if (value == true) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}
 
// Code contributed by Shivani Sanjay Shinde.

Python3

# Python 3program to check if a
# string is of the form a^nb^n.
 
# Returns true str is of the
# form a^nb^n.
 
 
def isAnBn(str):
 
    n = len(str)
 
    # After this loop 'i' has
    # count of a's
    for i in range(n):
        if (str[i] != 'a'):
            break
 
    # Since counts of a's and b's should
    # be equal, a should appear exactly
    # n/2 times
    if (i * 2 != n):
        return False
 
    # Rest of the characters must
    # be all 'b'
    for j in range(i, n):
        if (str[j] != 'b'):
            return False
 
    return True
 
 
# Driver code
if __name__ == "__main__":
    str = "abab"
    print("Yes") if isAnBn(str) else print("No")
 
# This code is contributed
# by ChitraNayal

C#

// C# program to check if a string
// is of the form a^ nb ^ n.
using System;
 
class GFG {
 
    // Function returns true str is of the form a^nb^n.
    public static bool isAnBn(String s)
    {
        int l = s.Length;
 
        // Only even length strings will have
        // same number of a's and b's
        if (l % 2 == 1) {
            return false;
        }
 
        // Set two pointers, one from the
        // left and another from right
        int i = 0;
        int j = l - 1;
 
        // Compare the characters
        // till the center
        while (i < j) {
            if (s[i] != 'a' || s[j] != 'b') {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
 
    // Driver Code
    public static void Main()
    {
        String s = "abab";
 
        // Function call
        bool value = isAnBn(s);
        if (value == true) {
            Console.Write("Yes");
        }
        else {
            Console.Write("No");
        }
    }
}
 
// This code is contributed by Nitin Mittal.

Javascript

// Javascript program to check if a string is of
// the form a^nb^n.
 
function isAnBn(s)
{
    let l = s.length;
  
        // Only even length strings will have same number of
        // a's and b's
        if (l % 2 == 1) {
            return false;
        }
        // Set two pointers, one from the left and another
        // from right
        let i = 0;
        let j = l - 1;
  
        // Compare the characters till the center
        while (i < j) {
            if (s[i] != 'a' || s[j] != 'b') {
                return false;
            }
            i++;
            j--;
        }
        return true;
}
 
let s = "abab";
 
// Function call
let value = isAnBn(s);
if (value == true) {
    document.write("Yes");
}
else {
    document.write("No");
}
 
 
// This code is contributed by ab2127

输出

No

另一种方法：
这个想法是从第一个和最后一个检查元素，如果在任何阶段我们的条件不满足，则返回 false。

下面是上述代码的实现：

C++

// C++ code to check a^nb^n
// pattern
#include 
using namespace std;
 
// Returns "Yes" str is of the form a^nb^n.
string isAnBn(string str)
{
    int n = str.length();
    if (n & 1)
        return "No";
     
    // check first half is 'a' and other half is full of 'b'
    int i;
    for (i = 0; i < n / 2; i++)
        if (str[i] != 'a' || str[n - i - 1] != 'b')
            return "No";
    return "Yes";
}
 
// Driver code
int main()
{
    string str = "ab";
   
    // Function call
    cout << isAnBn(str);
    return 0;
}

Java

// Java code to check a^nb^n
// pattern
import java.io.*;
class GFG {
 
  // Returns "Yes" str is of the form a^nb^n.
  static String isAnBn(String str)
  {
    int n = str.length();
    if ((n & 1) != 0)
      return "No";
 
    // check first half is 'a' and other half is full of 'b'
    int i;
    for (i = 0; i < n / 2; i++)
      if (str.charAt(i) != 'a' || str.charAt(n - i - 1) != 'b')
        return "No";
    return "Yes";
  }
 
  // Driver code
  public static void main (String[] args)
  {
    String str = "ab";
 
    // Function call
    System.out.println(isAnBn(str));
  }
}
 
// This code is contributed by rag2127

Python3

# Python3 code to check
# a^nb^n pattern
 
def isanbn(str):
  n=len(str)
   
  # if length of str is odd return No
  if n&1:
    return "No"
   
  # check first half is 'a' and other half is full of 'b'
  for i in range(int(n/2)):
    if str[i]!='a' or str[n-i-1]!='b':
      return "No"
  return "Yes"
     
# Driver code
input_str = "ab"
 
# Function call
print(isanbn(input_str))

C＃

// C# code to check a^nb^n
// pattern
using System;
public class GFG
{
 
  // Returns "Yes" str is of the form a^nb^n.
  static string isAnBn(string str)
  {
    int n = str.Length;
    if ((n & 1) != 0)
      return "No";
 
    // check first half is 'a' and other half is full of 'b'
    int i;
    for (i = 0; i < n / 2; i++)
      if (str[i] != 'a' || str[n-i-1] != 'b')
        return "No";
    return "Yes";
  }
 
  // Driver code
  static public void Main ()
  {
    string str = "ab";
 
    // Function call
    Console.WriteLine(isAnBn(str));
  }
}
 
// This code is contributed by avanitrachhadiya2155

Javascript

// Javascript code to check a^nb^n
// pattern
 
// Returns "Yes" str is of the form a^nb^n.
function isAnBn(str)
{
    let n = str.length;
    if ((n & 1) != 0)
      return "No";
  
    // check first half is 'a' and other half is full of 'b'
    let i;
    for (i = 0; i < n / 2; i++)
      if (str[i] != 'a' || str[n - i - 1] != 'b')
        return "No";
    return "Yes";
}
 
// Driver code
let str = "ab";
 
// Function call
document.write(isAnBn(str));
 
// This code is contributed by unknown2108

输出

Yes