[LeetCode]024. Swap Nodes in Pairs

Given a linked list, swap every two adjacent nodes and return its head.
For example, Given  1->2->3->4 , you should return the list as  2->1->4->3 .
Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.
Solution: draw a graph, 1->2->3->4, and then find the relationship.
Running time: O(n);
Note: Create the top node to reference the head; create the node pre to express the connection. 

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(head == null || head.next == null){
            return head;
        }
        ListNode cur = head;
        ListNode top = null;
        ListNode pre = null;
        while(cur.next != null){
            ListNode first = cur;
            ListNode second = first.next;
            first.next = second.next;
            second.next = first;
            // use a top reference point to the head node;
            if(top == null){
                top = second;
            }
            //use a pre reference to express the pre and after relationship: testcase: (1,2,3,4)
            if(pre == null){
                pre = first;
            }else{
                pre.next = second;
                pre = first;
            }
            if(first.next == null){
                break;
            }else{
                cur = first.next;
            }
        }
        return top;
    }
}