[GeeksForGeeks] Swap nodes in a single linked list by changing links

Given a singly linked list whose nodes contain an integer as their keys. All keys are distinct. 

Swap the node that has key x with the node that has key y.  

Nothing is done if either x or y does not exist in the given linked list. 

Do this swap by changing node links, not by swaping key values.

Key notes:

1. Use dummy node to simply the case that either x or y is the head node.

2.  if x and y are not adjacent, then there is 4 links that need to be changed;

　  if they are adjacent, then there is only 3 links that need to be changed; 

As a result, these 2 cases should be handled separately.

class ListNode{
    int key;
    ListNode next;
    ListNode(int key){
        this.key = key;
        this.next = null;
    }
}
public class Solution {
    public ListNode swapTwoNodesOfGivenKeys(ListNode head, int x, int y){
        if(head == null || head.next == null || x == y){
            return head;
        }
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        
        ListNode prevX = null, X = null, prevY = null, Y = null;
        ListNode prevNode = dummy, currNode = head;
        boolean foundX = false, foundY = false;
        
        while(currNode != null){
            if(currNode.key == x){
                prevX = prevNode;
                X = currNode;
                foundX = true;
            }
            else if(currNode.key == y){
                prevY = prevNode;
                Y = currNode;
                foundY = true;
            }
            if(foundX && foundY){
                break;
            }
            prevNode = currNode;
            currNode = currNode.next;
        }
        if(!foundX || !foundY){
            return dummy.next;
        }
        if(X == prevY){
            prevX.next = Y;
            X.next = Y.next;
            Y.next = X;
        }
        else if(Y == prevX){
            prevY.next = X;
            Y.next = X.next;
            X.next = Y;
        }
        else{
            prevX.next = Y;
            ListNode temp = Y.next;
            Y.next = X.next;
            X.next = temp;
            prevY.next = X;            
        }
        return dummy.next;
    }
}