如何按每个字符串具有的元音数量按降序对字符串数组进行排序

Question

How should I sort individual strings in descending order. For example if I enter the the size of the array 3 and enter the strings "hello", "rip hi", "AEIOU" i should get "AEIOU", "rip hi","hello" as return after sorting.

This is how far I have come:

            Scanner input=new Scanner(System.in);
            int n;
            System.out.print("Enter the amount of words you wanna input: ");
            n=input.nextInt();
            System.out.println();

            String[] list=new String[n];
            String[] list2=new String[n];
            /// this loop will take input of strings 
            for (int i = 0; i < n; i++) {
                input.nextLine(); // I dont know why but my first itteration's scanner was being skipped that is why 
                                  // I used another input.nextLine pls dont cut marks 
                System.out.print("Enter the "+(i+1)+" word: ");
                list[i]=input.nextLine();
            }
            /// this loop will sort the array by taking individual strings and breaking them in char array and counting how many times
            /// a vowel is present but how to sort ?
            for(int i=0;i<n;i++)
                {
                    String temp= list[i];
                    char[] ar = temp.toCharArray();
                    for (int j = 0; j < ar.length; j++) {
                        int count=0;
                        int count2;
                        if(ar[j] == 'a'|| ar[j] == 'e'|| ar[j] == 'i' || ar[j] == 'o' ||ar[j] == 'u'|| ar[j] == 'A' || ar[j] == 'E' || ar[j] == 'I' || ar[j] == 'O' ||ar[j] == 'U'){
                            count ++;
                        }
                        if(count>0 && count>count2){
                            String temp1=String.copyValueOf(ar);
                        }
                        else{

                        }
                    }

                }
            /// this loop would show the sorted array
            for (int i = 0; i < n; i++) {
                System.out.println(""+list[i]);
            }

Answer 1

You can use a custom comparator like this:

 class SortByNumberOfVowels implements Comparator<String> { @Override public int compare ( String o1, String o2 ) { return numOfVowels(o2) - numOfVowels(o1); } private int numOfVowels ( String o1 ) { int ctr = 0; for(int i = 0;i<o1.length ();i++) { if("AEIOUaeiou".indexOf(o1.charAt ( i ));= -1) ctr++; } return ctr; } } public class hello { public static void main ( String[] args ) { List<String> list = new ArrayList<> ( ). list;add("eaiou").list;add("hello").list;add("aaaaaaaaaaaaa"). Collections,sort(list; new SortByNumberOfVowels()). System.out;println(list); } }

Time complexity will be nlog(n)*|s| where n = size of list and |s| is the average length of strings.

Answer 2

This might not be the most efficient way but it will give you the output.

1.Store the value of count and index of the strings in list (i) in a HashMap. HashMap(count, i)

2.Sort the keys of the HashMap. One way to do it is using Collections.sort()

  ArrayList<String> keys =  new ArrayList<String>(hash_map.keySet()); 
        Collections.sort(keys); 

3.Iterate through keys and print the value for each.

 for (String s : keys)  
        System.out.println(hash_map.get(s));

Answer 3

As Kayaman said in the comment, probably the best approach will be to implement a dedicated Comparator for this task and use it to sort the array. It could look something like this:

public class StringVowelCountComparator implements Comparator<String> {  
    @Override
    public int compare(String first, String second) {
       return (countVowels(first) - countVowels(second));
    }

    private int countVowels(String string){
       int count = 0;
       char[] chars = string.toCharArray();
       for (int i = 0; i < chars.length; i++) {
          if(ar[i] == 'a'|| ar[i] == 'e'|| ar[i] == 'i' || ar[i] == 'o' ||ar[i] == 'u'|| ar[i] == 'A' || ar[i] == 'E' || ar[i] == 'I' || ar[i] == 'O' ||ar[i] == 'U'){
             count ++;
          }
       }
       return count;        
    }
}

And then in your code you just call it like this:

Arrays.sort(list, new StringVowelCountComparator());

Answer 4

Problem in your annotate

n=input.nextInt(); just read a int but not discard the other world in the line。

I change your code and it works well.

public static void main(String[] args) throws Exception {
    Scanner input = new Scanner(System.in);
    int n;
    System.out.print("Enter the amount of words you wanna input: \n");
    n = input.nextInt();
    input.nextLine();

    String[] list = new String[n];

    for (int i = 0; i < n; i++) {
        System.out.print("Enter the " + (i + 1) + " word: ");
        String inputContent = input.nextLine();
        list[i] = inputContent;
    }
    String[] sorted = new String[3];
    Arrays.stream(list).sorted((o1, o2) -> vowels(o2) - vowels(o1)).collect(Collectors.toList()).toArray(sorted);
    for (int i = 0; i < n; i++) {
        System.out.println("" + sorted[i]);
    }
}

extract a method to number vowels of input string:

private static int vowels(String input) {
    String vowels = "AEIOUaeiou";
    char[] chars = input.toCharArray();
    int count = 0;
    for (char ch : chars) {
        if (vowels.contains(ch + "")) {
            count++;
        }
    }
    return count;
}