如何将节点插入到具有给定数据的列表中并保持列表在Java中排序

Question

How do i add 100 to the linked list while keeping it sorted? I know i have to add an item to the front of a single-linked list and make it point to the first node of the list but am having much trouble in finding ways to start this problem off.

       IntegerNode n3 = new IntegerNode(9, null);
       IntegerNode n2 = new IntegerNode(5, n3);
       IntegerNode head = new IntegerNode(1, n2);

       IntegerNode curr;
       IntegerNode prev;

       //print all the items in the linked-list
       for(curr = head; curr!=null; curr = curr.getNext()) {
           System.out.println(curr.getItem());
       }

       int data = 100;

       //insert an node to the list with the given data, and maintain the list to be sorted
    }
}

Answer 1

The following insert method should be a start:

public void insert(int val){
    if(this.head==null){ //empty
        this.head = new IntegerNode (val);
        this.head.setNext(null);
        return;
    }
    IntegerNode  prev = null;
    IntegerNode  curr = this.head;
    IntegerNode  newNode = new SLLNode(val);
    while(curr!=null && curr.getVal()<val){
        prev = curr;
        curr = curr.getNext();
    }
    if(prev==null){//new element to be placed at first
        newNode.setNext(curr);
        this.head = newNode;
    }
    else if(curr==null){// new element to be placed at end
        newNode.setNext(null);
        prev.setNext(newNode);
    }else{ //intermediate position
        newNode.setNext(curr);
        prev.setNext(newNode);
    }
}

Steps are as follows:

1. If the head is null, make a node with the current value and set it as root.
2. Else iterate through the list to find the largest node with value less than the current node.We keep references to the previous and next nodes of this node.
3. If the prev is null, that menas the value we are inserting is smaller than all values currently in list and should be placed at the first.
4. If the curr is null, that means we have reached the end of the list iterating and our value is larger than all the nodes currently in list.So we set the node at last.

Hope this helps.