检查给定的字符串是否是反向双音字符串

给定字符串str ，任务是检查该字符串是否为反向双音字符串。如果字符串str是反向Bitonic字符串，则打印“ YES” 。否则，打印“否” 。

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

例子：

Input: str = “zyxbcd” 
Output: YES 
Explanation: 
In the above string, the ASCII values first decreases {z, y, x} and then increases {b, c, d}.

Input: str = “abcdwef” 
Output: NO 

方法：
为了解决这个问题，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。
• 严格减少，然后严格增加。

请按照以下步骤解决问题：

1. 遍历字符串，对于每个字符，检查下一个字符的ASCII值是否小于当前字符的ASCII值。
2. 如果在任何时候下一个字符的ASCII值大于当前字符的ASCII值，请中断循环。
3. 现在从该索引开始遍历，对于每个字符，检查下一个字符的ASCII值是否大于当前字符的ASCII值。
4. 如果在到达数组末尾之前的任何时候，下一个字符的ASCII值小于当前字符的ASCII值，则打印“ NO”并中断循环。
5. 如果成功遍历了整个字符串，请打印“ YES” 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if the given
// string is reverse bitonic
int checkReverseBitonic(string s)
{
    int i, j;
 
    // Check for decreasing 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 increasing 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;
 
    // If the string is
    // reverse bitonic
    return 1;
}
 
// Driver Code
int main()
{
    string s = "abcdwef";
 
    (checkReverseBitonic(s) == 1)
        ? cout << "YES"
        : cout << "NO";
 
    return 0;
}

Java

// Java program to implement
// the above approach
class GFG{
     
// Function to check if the given
// string is reverse bitonic
static int checkReverseBitonic(String s)
{
    int i, j;
 
    // Check for decreasing sequence
    for(i = 1; i < s.length(); i++)
    {
       if (s.charAt(i) < s.charAt(i - 1))
           continue;
        
       if (s.charAt(i) >= s.charAt(i - 1))
           break;
    }
     
    // If end of string has
    // been reached
    if (i == s.length() - 1)
        return 1;
 
    // Check for increasing sequence
    for(j = i + 1; j < s.length(); j++)
    {
       if (s.charAt(j) > s.charAt(j - 1))
           continue;
       if (s.charAt(j) <= s.charAt(j - 1))
           break;
    }
    i = j;
     
    // If the end of string
    // hasn't been reached
    if (i != s.length())
        return 0;
 
    // If the string is
    // reverse bitonic
    return 1;
}
 
// Driver Code
public static void main(String []args)
{
    String s = "abcdwef";
 
    if(checkReverseBitonic(s) == 1)
        System.out.println("YES");
    else
        System.out.println("NO");
}
}
 
// This code is contributed by grand_master

Python3

# Python3 program to implement
# the above approach
 
# Function to check if the given
# string is reverse bitonic
def checkReverseBitonic(s):
  
    i = 0
    j = 0
 
    # Check for decreasing sequence
    for  i in range(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 1;
 
    # Check for increasing 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)) :
        return 0;
 
    # If reverse bitonic
    return 1;
 
  
# Given string
s = "abcdwef"
# Function Call
if(checkReverseBitonic(s) == 1) :
    print("YES")
else:
    print("NO")

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
     
// Function to check if the given
// string is reverse bitonic
static int checkReverseBitonic(String s)
{
    int i, j;
 
    // Check for decreasing 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 increasing 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;
 
    // If the string is
    // reverse bitonic
    return 1;
}
 
// Driver Code
public static void Main(String []args)
{
    String s = "abcdwef";
 
    if(checkReverseBitonic(s) == 1)
        Console.WriteLine("YES");
    else
        Console.WriteLine("NO");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to check if the given
// string is reverse bitonic
function checkReverseBitonic(s)
{
    var i, j;
 
    // Check for decreasing 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 increasing 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;
 
    // If the string is
    // reverse bitonic
    return 1;
}
 
// Driver Code
var s = "abcdwef";
(checkReverseBitonic(s) == 1) ?
document.write("YES") :
document.write("NO");
     
// This code is contributed by rutvik_56
     

输出：

NO

时间复杂度： O(N)
辅助空间： O(1)

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