双音字符串

给定字符串str ，任务是检查该字符串是否是Bitonic字符串。如果字符串str是Bitonic String，则输出“ YES”，否则输出“ NO” 。

A Bitonic String is a string in which the characters are arranged in increasing order followed by decreasing order of their ASCII values. 
 

例子：

Input: str = “abcdfgcba” 
Output: YES 
Explanation: 
In the above string, the ASCII values first increases {a, b, c, d, f, g} and then decreases {g, c, b, a}.
Input: str = “abcdwef” 
Output: NO 
 

方法：
为了解决这个问题，我们只需要遍历字符串并检查字符串的字符的ASCII值以下任何模式：

• 严格增加。
• 严格减少。
• 严格增加，然后严格减少。

请按照以下步骤解决问题：

1. 开始遍历字符串，并继续检查下一个字符的ASCII值是否大于当前字符的ASCII值。
2. 如果在任何时候下一个字符的ASCII值不大于当前字符的ASCII值，请中断循环。
3. 再次从上述中断索引开始遍历，并检查下一个字符的ASCII值是否小于当前字符的ASCII值。
4. 如果在到达数组末尾之前，下一个字符的ASCII值在任何时候都不小于当前字符的ASCII值，则打印“ NO”并中断循环。
5. 如果成功到达数组末尾，请打印“ YES” 。

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 the given
// string is bitonic
int checkBitonic(string s)
{
    int i, j;
  
    // Check for increasing sequence
    for (i = 1; i < s.size(); i++) {
        if (s[i] > s[i - 1])
            continue;
  
        if (s[i] <= s[i - 1])
            break;
    }
  
    // If end of string has been reached
    if (i == s.size() - 1)
        return 1;
  
    // Check for decreasing sequence
    for (j = i + 1; j < s.size();
         j++) {
        if (s[j] < s[j - 1])
            continue;
  
        if (s[j] >= s[j - 1])
            break;
    }
  
    i = j;
  
    // If the end of string hasn't
    // been reached
    if (i != s.size())
        return 0;
  
    // Return true if bitonic
    return 1;
}
  
// Driver Code
int main()
{
    // Given string
    string s = "abcdfgcba";
  
    // Function Call
    (checkBitonic(s) == 1)
        ? cout << "YES"
        : cout << "NO";
  
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
  
class GFG{
  
// Function to check if the given
// String is bitonic
static int checkBitonic(char []s)
{
    int i, j;
  
    // Check for increasing sequence
    for(i = 1; i < s.length; i++)
    {
        if (s[i] > s[i - 1])
            continue;
  
        if (s[i] <= s[i - 1])
            break;
    }
  
    // If end of String has been reached
    if (i == s.length - 1)
        return 1;
  
    // Check for decreasing sequence
    for(j = i + 1; j < s.length; j++)
    {
        if (s[j] < s[j - 1])
            continue;
  
        if (s[j] >= s[j - 1])
            break;
    }
    i = j;
  
    // If the end of String hasn't
    // been reached
    if (i != s.length)
        return 0;
  
    // Return true if bitonic
    return 1;
}
  
// Driver Code
public static void main(String[] args)
{
  
    // Given String
    String s = "abcdfgcba";
  
    // Function Call
    System.out.print((checkBitonic(
        s.toCharArray()) == 1) ? "YES" : "NO");
}
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python3 program for the above approach 
  
# Function to check if the given 
# string is bitonic 
def checkBitonic(s): 
      
    i = 0
    j = 0
  
    # Check for increasing sequence 
    for i in range(1, len(s)): 
        if (s[i] > s[i - 1]):
            continue; 
  
        if (s[i] <= s[i - 1]):
            break; 
      
    # If end of string has been reached 
    if (i == (len(s) - 1)):
        return True;
      
    # Check for decreasing sequence 
    for j in range(i + 1, len(s)):
        if (s[j] < s[j - 1]):
            continue;
              
        if (s[j] >= s[j - 1]):
            break;
    i = j; 
  
    # If the end of string hasn't 
    # been reached
    if (i != len(s) - 1):
        return False; 
  
    # Return true if bitonic
    return True; 
  
# Driver code
  
# Given string
s = "abcdfgcba"
  
# Function Call
if(checkBitonic(s)):
    print("YES")
else:
    print("NO") 
      
# This code is contributed by grand_master

C#

// C# program for the above approach 
using System;
  
class GFG{ 
  
// Function to check if the given 
// String is bitonic 
static int checkBitonic(char []s) 
{ 
    int i, j; 
  
    // Check for increasing sequence 
    for(i = 1; i < s.Length; i++) 
    { 
        if (s[i] > s[i - 1]) 
            continue; 
  
        if (s[i] <= s[i - 1]) 
            break; 
    } 
  
    // If end of String has been reached 
    if (i == s.Length - 1) 
        return 1; 
  
    // Check for decreasing sequence 
    for(j = i + 1; j < s.Length; j++) 
    { 
        if (s[j] < s[j - 1]) 
            continue; 
  
        if (s[j] >= s[j - 1]) 
            break; 
    } 
    i = j; 
  
    // If the end of String hasn't 
    // been reached 
    if (i != s.Length) 
        return 0; 
  
    // Return true if bitonic 
    return 1; 
} 
  
// Driver Code 
public static void Main(String[] args) 
{ 
  
    // Given String 
    String s = "abcdfgcba"; 
  
    // Function call 
    Console.Write((checkBitonic( 
        s.ToCharArray()) == 1) ? "YES" : "NO"); 
} 
} 
  
// This code is contributed by PrinciRaj1992

输出：

YES

时间复杂度： O(N)
辅助空间： O(1)

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