根据长度升序对字符串数组进行排序

Question

The question is to print the array of string in accordance with the length of strings in ascending order.

For example

input={"vellore","i","from","am"}
output=i am from vellore

Here is my code:

int n=sc.nextInt();
    sc.nextLine();
    String[] arr = new String[n];
    for(int i=0;i<n;i++){
        arr[i]=sc.nextLine();
    }
    //System.out.println(Arrays.toString(arr));
    Arrays.sort(arr);
    for(String i: arr){
        System.out.print(i+" ");
    }

Now I know that my output will come in lexical order that is as "am from i vellore", but I want to get my desired output using sort method. I tried using Collections.sort() as well by using arraylist but I still didn't get my desired output.

I want to get my output using sort method without using the normal approach by comparing the lengths of string and all.

Answer 1

You want a Comparator . In this case, you specifically want to start by comparing the String lengths. I would suggest you then compare naturally (to break ties). Like,

String[] arr = { "vellore", "i", "from", "am" };
Arrays.sort(arr, Comparator.comparingInt(String::length)
        .thenComparing(Comparator.naturalOrder()));
System.out.println(Arrays.toString(arr));

Outputs (as requested)

[i, am, from, vellore]

Answer 2

tl;dr

Comparator
.comparingLong( ( String s ) -> s.codePoints().count() )
.thenComparing( Comparator.naturalOrder() )

Avoid legacy type char

The Answer by Frisch is correct in suggesting the use of a Comparator . However, the method reference seen there, String::length , fails with most characters. See this example . The String#length method reports a two-character string like c as three, incorrectly.

The failure is because that method depends on char which has been legacy since Java 2. As a 16-bit value, the char type is physically incapable of representing most of the over 140,000 characters defined in a Unicode.

Code points

Instead, use Unicode code point integers.

Switch out that method reference for the use of code points rather than char type.

Here we use String#codePoints to generate an IntStream . That IntStream is a stream of each character's code point number, a value in the range of zero to just over a million. Then we use IntStream#count to get a count of the values in that stream, a count of the characters in the original string.

    String[] arr = { "a1", "c🕒", "b2", "d4" };
    Arrays
        .sort(
            arr , 
            Comparator
            .comparingLong( ( String s ) -> s.codePoints().count() )
            .thenComparing( Comparator.naturalOrder() )
        );
    System.out.println( Arrays.toString( arr ) );

See that code run live at IdeOne.com .

[a1, b2, c, d4]