给定一个非空字符串S ，任务是从字符串S中打印出最长的子序列，其中包含交替的元音和辅音。
注意：如果存在多个具有相同长度的子序列，请打印其字符的ASCII值最大的子序列。
例子：
Input: S = “geeksforgeeks”
Output: gesores
Explanation: “gekorek”, “gekores”, “gekogek”, “gekoges”, “gesorek”, “gesores”, “gesogek”, “gesoges”, “geforek”, “gefores”, “gefogek” and “gefoges” are the possible longest subsequences with alternating consonants and vowels. “gesores” is the subsequence with maximum sum of ASCII values of characters and hence it is the solution.
Input: S = “ababababab”
Output: ababababab
Explanation: “ababababab” is the longest possible subsequence containing alternating vowels and consonants.
方法：
请按照以下步骤解决问题：
- 将S的第一个字符的ASCII值存储为当前块的最大值( maxi )，并在标志中存储该字符的类型。如果字符是辅音，则将标志设置为0 ，否则设置为1 。
- 遍历字符串的其余部分。
- 对于每一个字符，检查它是否属于相同的块作为前一个字符与否。
- 如果它属于同一块，则将maxi更新为max(maxi，第i个字符的ASCII值)。
- 否则，在答案中附加具有ASCII值maxi的字符。将当前第i个字符的ASCII值存储为maxi 。更新标志(标志+ 1)％2来表示当前字符的类型。
- 遍历整个字符串，将具有ASCII值maxi的字符添加到答案中。打印代表子序列的最后一个字符串。
下面的代码是上述方法的实现：
C++
// C++ program to find the longest
// subsequence containing alternating
// vowels and consonants
#include
using namespace std;
// Function that return 1 or 0
// if the given character is
// vowel or consonant respectively
int isVowel(char c)
{
// Returns 1 if c is vowel
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return 1;
// Returns 0 if
// c is consonant
return 0;
}
// Function to find the longest
// subsequence containing
// alternate vowels and
// consonants
string Subsequence(string str)
{
// Store the length
// of the string
int n = str.length();
// Assume first character
// to be consonant
int flag = 0;
// If first character is vowel
if (isVowel(str[0]) == 1)
flag = 1;
// Store the ASCII value
int maxi = (int)str[0];
// Stores the final answer
string ans = "";
for(int i = 1; i < n; i++)
{
// If the current character belongs
// to same category (Vowel / Consonant)
// as the previous character
if (isVowel(str[i]) == flag)
{
// Store the maximum ASCII value
maxi = max(maxi, (int)str[i]);
}
// Otherwise
else
{
// Store the character with
// maximum ASCII value from
// the previous block
ans += (char)(maxi);
// Store the ASCII of the
// current character as the
// maximum of the current block
maxi = (int)str[i];
// Switch the type of the
// current block
flag = (flag + 1) % 2;
}
}
// Store the chracter with
// maximum ASCII value
// from the last block
ans += (char)(maxi);
// Return the result
return ans;
}
// Driver code
int main()
{
string str = "geeksforgeeks";
cout << (Subsequence(str));
}
// This code is contributed by chitranayal Java
// Java program to find the longest
// subsequence containing alternating
// vowels and consonants
import java.util.*;
import java.lang.*;
import java.io.*;
import java.math.*;
class GFG {
// Function that return 1 or 0
// if the given character is
// vowel or consonant respectively
static int isVowel(char c)
{
// Returns 1 if c is vowel
if (c == 'a' || c == 'e'
|| c == 'i' || c == 'o'
|| c == 'u')
return 1;
// Returns 0 if
// c is consonant
return 0;
}
// Function to find the longest
// subsequence containing
// alternate vowels and
// consonants
static String Subsequence(String str)
{
// Store the length
// of the string
int n = str.length();
// Assume first character
// to be consonant
int flag = 0;
// If first character is vowel
if (isVowel(str.charAt(0)) == 1)
flag = 1;
// Store the ASCII value
int maxi = (int)str.charAt(0);
// Stores the final answer
String ans = "";
for (int i = 1; i < n; i++) {
// If the current character belongs
// to same category (Vowel / Consonant)
// as the previous character
if (isVowel(str.charAt(i)) == flag) {
// Store the maximum ASCII value
maxi = Math.max(maxi,
(int)str.charAt(i));
}
// Otherwise
else {
// Store the character with
// maximum ASCII value from
// the previous block
ans += (char)(maxi);
// Store the ASCII of the
// current character as the
// maximum of the current block
maxi = (int)str.charAt(i);
// Switch the type of the
// current block
flag = (flag + 1) % 2;
}
}
// Store the chracter with
// maximum ASCII value
// from the last block
ans += (char)(maxi);
// Return the result
return ans;
}
// Driver Program
public static void main(String[] args)
{
String str = "geeksforgeeks";
System.out.println(Subsequence(str));
}
}Python3
# Python3 program to find the longest
# subsequence containing alternating
# vowels and consonants
def isVowel(c):
# boolean function that check whether
# the given char is vowel or not
# and returns a boolean value respectively
vowels=['a','e','i','o','u']
if(c in vowels):
return True
return False
def Subsequence(str):
#string that stores the final result
ans=''
flag=(isVowel(str[0]))
#taking the first character
#as the maximum ASCII valued char
maxi=ord(str[0])
for i in range(1,len(str)):
# If the current character belongs to
# same category(Vowel / Consonant) as the
# previous character
if (isVowel(str[i]) == flag):
# choosing a maximum ASCII valued char
maxi=max(maxi,ord(str[i]))
#otherwise
else:
ans=ans+chr(maxi)
maxi=ord(str[i])
#toggling the flag
flag=not(flag)
#adding the last char to the answer
ans=ans+chr(maxi)
return ans
#Driver program
if __name__ == "__main__":
input_string ='geeksforgeeks'
print(Subsequence(input_string))
# Contributed by
# Nvss Maneesh GuptaC#
// C# program to find the longest
// subsequence containing alternating
// vowels and consonants
using System;
class GFG{
// Function that return 1 or 0
// if the given character is
// vowel or consonant respectively
static int isVowel(char c)
{
// Returns 1 if c is vowel
if (c == 'a' || c == 'e' ||
c == 'i' || c == 'o' ||
c == 'u')
return 1;
// Returns 0 if
// c is consonant
return 0;
}
// Function to find the longest
// subsequence containing
// alternate vowels and
// consonants
static String Subsequence(String str)
{
// Store the length
// of the string
int n = str.Length;
// Assume first character
// to be consonant
int flag = 0;
// If first character is vowel
if (isVowel(str[0]) == 1)
flag = 1;
// Store the ASCII value
int maxi = (int)str[0];
// Stores the readonly answer
String ans = "";
for(int i = 1; i < n; i++)
{
// If the current character belongs
// to same category (Vowel / Consonant)
// as the previous character
if (isVowel(str[i]) == flag)
{
// Store the maximum ASCII value
maxi = Math.Max(maxi, (int)str[i]);
}
// Otherwise
else
{
// Store the character with
// maximum ASCII value from
// the previous block
ans += (char)(maxi);
// Store the ASCII of the
// current character as the
// maximum of the current block
maxi = (int)str[i];
// Switch the type of the
// current block
flag = (flag + 1) % 2;
}
}
// Store the chracter with
// maximum ASCII value
// from the last block
ans += (char)(maxi);
// Return the result
return ans;
}
// Driver code
public static void Main(String[] args)
{
String str = "geeksforgeeks";
Console.WriteLine(Subsequence(str));
}
}
// This code is contributed by 29AjayKumar输出
gesores
时间复杂度： O(N)