Java how to sort a Linked List?

Question

I am needing to sort a linked list alphabetically. I have a Linked List full of passengers names and need the passengers name to be sorted alphabetically. How would one do this? Anyone have any references or videos?

Answer 1

You can use Collections#sort to sort things alphabetically.

Answer 2

In order to sort Strings alphabetically you will need to use a Collator, like:

 LinkedList<String> list = new LinkedList<String>();
 list.add("abc");
 list.add("Bcd");
 list.add("aAb");
 Collections.sort(list, new Comparator<String>() {
     @Override
     public int compare(String o1, String o2) {
         return Collator.getInstance().compare(o1, o2);
     }
 });

Because if you just call Collections.sort(list) you will have trouble with strings that contain uppercase characters.

For instance in the code I pasted, after the sorting the list will be: [aAb, abc, Bcd] but if you just call Collections.sort(list); you will get: [Bcd, aAb, abc]

Note: When using a Collator you can specify the locale Collator.getInstance(Locale.ENGLISH) this is usually pretty handy.

Answer 3

In java8 you no longer need to use Collections.sort method as LinkedList inherits the method sort from java.util.List, so adapting Fido's answer to Java8:

    LinkedList<String>list = new LinkedList<String>();
    list.add("abc");
    list.add("Bcd");
    list.add("aAb");

    list.sort( new Comparator<String>(){
    @Override
        public int compare(String o1,String o2){
            return Collator.getInstance().compare(o1,o2);
        }
    });

References:

http://docs.oracle.com/javase/8/docs/api/java/util/LinkedList.html

http://docs.oracle.com/javase/7/docs/api/java/util/List.html

Answer 4

Elegant solution since JAVA 8:

LinkedList<String>list = new LinkedList<String>();
list.add("abc");
list.add("Bcd");
list.add("aAb");

list.sort(String::compareToIgnoreCase);

---

Another option would be using lambda expressions:

list.sort((o1, o2) -> o1.compareToIgnoreCase(o2));

Answer 5

Here is the example to sort implemented linked list in java without using any standard java libraries.

package SelFrDemo;

class NodeeSort {
    Object value;
    NodeeSort next;

    NodeeSort(Object val) {
        value = val;
        next = null;

    }

    public Object getValue() {
        return value;
    }

    public void setValue(Object value) {
        this.value = value;
    }

    public NodeeSort getNext() {
        return next;
    }

    public void setNext(NodeeSort next) {
        this.next = next;
    }

}

public class SortLinkList {
    NodeeSort head;
    int size = 0;

    NodeeSort add(Object val) {
        // TODO Auto-generated method stub
        if (head == null) {
            NodeeSort nodee = new NodeeSort(val);
            head = nodee;
            size++;
            return head;
        }
        NodeeSort temp = head;

        while (temp.next != null) {
            temp = temp.next;
        }

        NodeeSort newNode = new NodeeSort(val);
        temp.setNext(newNode);
        newNode.setNext(null);
        size++;
        return head;
    }

    NodeeSort sort(NodeeSort nodeSort) {

        for (int i = size - 1; i >= 1; i--) {
            NodeeSort finalval = nodeSort;
            NodeeSort tempNode = nodeSort;

            for (int j = 0; j < i; j++) {

                int val1 = (int) nodeSort.value;
                NodeeSort nextnode = nodeSort.next;
                int val2 = (int) nextnode.value;
                if (val1 > val2) {

                    if (nodeSort.next.next != null) {
                        NodeeSort CurrentNext = nodeSort.next.next;
                        nextnode.next = nodeSort;
                        nextnode.next.next = CurrentNext;
                        if (j == 0) {
                            finalval = nextnode;
                        } else
                            nodeSort = nextnode;

                        for (int l = 1; l < j; l++) {
                            tempNode = tempNode.next;
                        }

                        if (j != 0) {
                            tempNode.next = nextnode;

                            nodeSort = tempNode;
                        }
                    } else if (nodeSort.next.next == null) {
                        nextnode.next = nodeSort;
                        nextnode.next.next = null;
                        for (int l = 1; l < j; l++) {
                            tempNode = tempNode.next;
                        }
                        tempNode.next = nextnode;
                        nextnode = tempNode;
                        nodeSort = tempNode;

                    }

                } else
                    nodeSort = tempNode;
                nodeSort = finalval;
                tempNode = nodeSort;
                for (int k = 0; k <= j && j < i - 1; k++) {
                    nodeSort = nodeSort.next;
                }

            }

        }
        return nodeSort;

    }

    public static void main(String[] args) {
        SortLinkList objsort = new SortLinkList();
        NodeeSort nl1 = objsort.add(9);
        NodeeSort nl2 = objsort.add(71);
        NodeeSort nl3 = objsort.add(6);
        NodeeSort nl4 = objsort.add(81);
        NodeeSort nl5 = objsort.add(2);

        NodeeSort NodeSort = nl5;

        NodeeSort finalsort = objsort.sort(NodeSort);
        while (finalsort != null) {
            System.out.println(finalsort.getValue());
            finalsort = finalsort.getNext();
        }

    }
}

Answer 6

Node mergeSort(Node head) {
    if(head == null || head.next == null) {
        return head;
    }

    Node middle = middleElement(head);
    Node nextofMiddle = middle.next;
    middle.next = null;

    Node left = mergeSort(head);
    Node right = mergeSort(nextofMiddle);

    Node sortdList = merge(left, right);

    return sortdList;
}

Node merge(Node left, Node right) {
    if(left == null) {
        return right;
    }

    if(right == null) {
        return left;
    }
    Node temp = null;
    if(left.data < right.data) {
        temp = left;
        temp.next = merge(left.next, right);
    } else {
        temp = right;
        temp.next = merge(left, right.next);
    }

    return temp;
}

Node middleElement(Node head) {
    Node slow = head;
    Node fast = head;
    while (fast != null && fast.next != null && fast.next.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    return slow;
}

Answer 7

You can do it by Java 8 lambda expression :

LinkedList<String> list=new LinkedList<String>();
list.add("bgh");
list.add("asd");
list.add("new");
//lambda expression
list.sort((a,b)->a.compareTo(b));

Answer 8

I wouldn't. I would use an ArrayList or a sorted collection with a Comparator. Sorting a LinkedList is about the most inefficient procedure I can think of.

Answer 9

If you'd like to know how to sort a linked list without using standard Java libraries, I'd suggest looking at different algorithms yourself. Examples here show how to implement an insertion sort, another StackOverflow post shows a merge sort, and ehow even gives some examples on how to create a custom compare function in case you want to further customize your sort.