83. Remove Duplicates from Sorted List

原题链接：https://leetcode.com/problems/remove-duplicates-from-sorted-list/description/
这道题目级别为 Easy，做起来也确实非常 easy。

/**
 * Created by clearbug on 2018/2/26.
 */
public class Solution {

    static class ListNode {
        int val;
        ListNode next;
        ListNode(int x) {
            val = x;
        }
    }

    public static void main(String[] args) {
        Solution s = new Solution();

        // test 1
        ListNode node1 = new ListNode(1);
        ListNode node2 = new ListNode(1);
        ListNode node3 = new ListNode(2);
        node1.next = node2;
        node2.next = node3;
        s.printList(node1);
        s.printList(s.deleteDuplicates(node1));

        // test 2
        ListNode node21 = new ListNode(1);
        ListNode node22 = new ListNode(1);
        ListNode node23 = new ListNode(2);
        ListNode node24 = new ListNode(3);
        ListNode node25 = new ListNode(3);
        node21.next = node22;
        node22.next = node23;
        node23.next = node24;
        node24.next = node25;
        s.printList(node21);
        s.printList(s.deleteDuplicates(node21));

        // test 3
        ListNode node31 = new ListNode(1);
        ListNode node32 = new ListNode(2);
        node31.next = node32;
        s.printList(node31);
        s.printList(s.deleteDuplicates(node31));
    }

    public ListNode deleteDuplicates(ListNode head) {
        ListNode currentNode = head;
        ListNode preNode = head;
        while (currentNode != null) {
            if (currentNode.val == preNode.val) {
                currentNode = currentNode.next;
            } else {
                preNode.next = currentNode;
                preNode = currentNode;
                currentNode = currentNode.next;
            }
        }

        if (preNode != null) {
            preNode.next = null;
        }
        return head;
    }

    private void printList(ListNode head) {
        while (head != null) {
            System.out.print(head.val + ", ");
            head = head.next;
        }
        System.out.println();
    }

}