用字符串中按字典顺序排列的下一个元音替换每个元音

给定大小为N 的字符串str ，其中包含小写英文字母。任务是按字典顺序用下一个直接元音替换每个元音，即，

‘a’ will be replaced by ‘e’, 
‘e’ will be replaced by ‘i’, 
‘i’ will be replaced by ‘o’, 
‘o’ will be replaced by ‘u’, 
‘u’ will be replaced by ‘a’.

例子：

Input: str = “geeksforgeeks” 
Output: giiksfurgiiks 
Explaination: 
e is replace by i 
o is replace by u 
So final string will be “giiksfurgiiks”.

Input: str = “gfg” 
Output: gfg 

方法：我们将创建一个大小为 5 的 Hashing 来存储所有元音，以便可以轻松地对每个元音进行替换。

• 创建一个映射并存储所有元音。
• 从左到右迭代字符串元素。
• 如果字符串元素是元音，则将其更改为下一个元音。
• 最后，打印最终的字符串。

下面是上述方法的实现：

C++14

// C++ program to convert all the vowels in
// in the string to the next vowel
 
#include 
using namespace std;
 
// Function to replace every vowel
// with next vowel lexicographically
string print_next_vovel_string(string str)
{
 
    // Storing the vowels in the map with
    // custom numbers showing  their index
    map m;
    m['a'] = 0;
    m['e'] = 1;
    m['i'] = 2;
    m['o'] = 3;
    m['u'] = 4;
    char arr[5] = { 'a', 'e', 'i', 'o', 'u' };
 
    int N = str.length();
 
    // Iterate over the string
    for (int i = 0; i < N; i++) {
 
        char c = str[i];
 
        // If the current character is a vowel
        // Find the index in Hash and
        // Replace it with next vowel from Hash
        if (c == 'a' || c == 'e' || c == 'i'
            || c == 'o' || c == 'u') {
 
            int index = m + 1;
            int newindex = index % 5;
            str[i] = arr[newindex];
        }
    }
 
    return str;
}
 
// Driver function
int main()
{
    string str = "geeksforgeeks";
 
    cout << print_next_vovel_string(str);
    return 0;
}

Java

// Java program to convert all the vowels in
// in the String to the next vowel
import java.util.*;
 
class GFG{
 
// Function to replace every vowel
// with next vowel lexicographically
static String print_next_vovel_String(char []str)
{
 
    // storing the vowels in the map with
    // custom numbers showing their index
    HashMap m = new HashMap();
    m.put('a', 0);
    m.put('e', 1);
    m.put('i', 2);
    m.put('o', 3);
    m.put('u', 4);
 
    char arr[] = { 'a', 'e', 'i', 'o', 'u' };
 
    int N = str.length;
 
    // Iterate over the String
    for(int i = 0; i < N; i++)
    {
        char c = str[i];
 
        // If the current character is a vowel
        // Find the index in Hash and
        // Replace it with next vowel from Hash
        if (c == 'a' || c == 'e' ||
            c == 'i' || c == 'o' ||
            c == 'u')
        {
            int index = m.get(c) + 1;
            int newindex = index % 5;
             
            str[i] = arr[newindex];
        }
    }
    return String.valueOf(str);
}
 
// Driver code
public static void main(String[] args)
{
    String str = "geeksforgeeks";
 
    System.out.print(print_next_vovel_String(
        str.toCharArray()));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program to convert
# all the vowels in the string
# to the next vowel
 
# Function to replace every vowel
# with next vowel lexicographically
def print_next_vovel_string(st):
 
    # Storing the vowels in
    # the map with custom
    # numbers showing  their index
    m = {}
    m['a'] = 0
    m['e'] = 1
    m['i'] = 2
    m['o'] = 3
    m['u'] = 4
    arr = ['a', 'e', 'i', 'o', 'u']
 
    N = len(st)
 
    # Iterate over the string
    for i in range (N):
 
        c = st[i]
 
        # If the current character
        # is a vowel
        # Find the index in Hash
        # and Replace it with next
        # vowel from Hash
        if (c == 'a' or c == 'e' or
            c == 'i'or c == 'o' or
            c == 'u'):           
            index = m[st[i]] + 1
            newindex = index % 5
            st = st.replace(st[i], arr[newindex], 1)
 
    return st
 
# Driver function
if __name__ == "__main__":
    st = "geeksforgeeks"
    print (print_next_vovel_string(st))
    
# This code is contributed by Chitranayal

C#

// C# program to convert all the vowels in
// in the String to the next vowel
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to replace every vowel
// with next vowel lexicographically
static String print_next_vovel_String(char []str)
{
   
    // Storing the vowels in the map with
    // custom numbers showing their index
    Dictionary m = new Dictionary();
    m.Add('a', 0);
    m.Add('e', 1);
    m.Add('i', 2);
    m.Add('o', 3);
    m.Add('u', 4);
 
    char []arr = { 'a', 'e', 'i', 'o', 'u' };
 
    int N = str.Length;
 
    // Iterate over the String
    for(int i = 0; i < N; i++)
    {
        char c = str[i];
 
        // If the current character is a vowel
        // Find the index in Hash and
        // Replace it with next vowel from Hash
        if (c == 'a' || c == 'e' ||
            c == 'i' || c == 'o' ||
            c == 'u')
        {
            int index = m + 1;
            int newindex = index % 5;
             
            str[i] = arr[newindex];
        }
    }
    return String.Join("", str);
}
 
// Driver code
public static void Main(String[] args)
{
    String str = "geeksforgeeks";
 
    Console.Write(print_next_vovel_String(
        str.ToCharArray()));
}
}
 
// This code is contributed by Amit Katiyar

Javascript

      // JavaScript program to convert
      // all the vowels in the string
      // to the next vowel
      // Function to replace every vowel
      // with next vowel lexicographically
      function print_next_vovel_string(st) {
        // Storing the vowels in
        // the map with custom
        // numbers showing their index
        var m = {};
        m["a"] = 0;
        m["e"] = 1;
        m["i"] = 2;
        m["o"] = 3;
        m["u"] = 4;
        var arr = ["a", "e", "i", "o", "u"];
 
        var N = st.length;
 
        // Iterate over the string
        for (let i = 0; i < N; i++) {
          var c = st[i];
 
          // If the current character
          // is a vowel
          // Find the index in Hash
          // and Replace it with next
          // vowel from Hash
          if (c === "a" || c === "e" || c === "i" || c === "o" || c === "u") {
            index = m[st[i]] + 1;
            newindex = index % 5;
            st = st.replace(st[i], arr[newindex]);
          }
        }
        return st;
      }
 
      // Driver function
      var st = "geeksforgeeks";
      document.write(print_next_vovel_string(st));
    

输出：

giiksfurgiiks

时间复杂度： O (N)
辅助空间： O (1)

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