leetcode023--删除排序链表中的重复元素

两种解法：

第一种思路：采用常规方法，判断当前节点的值是否和下一个节点相等，若相等，则让让当前节点的next域为下一个节点的下一个节点；

第二种思路： 采用递归的思想，如下：

//递归处理
    private static ListNode deleteDuplicates2(ListNode head){
        if (head == null || head.next == null){
            return head;
        }
        head.next = deleteDuplicates2(head.next);
        return head.val == head.next.val ? head.next : head;
    }

全部代码如下：

public class test29 {
    /**
     * 删除排序链表中的重复元素:  给定一个已排序的链表的头 head ， 删除所有重复的元素，使每个元素只出现一次 。返回已排序的链表 。
     */
    static class ListNode {
        int val;
        ListNode next;
        ListNode() {}
        ListNode(int val) { this.val = val; }
        ListNode(int val, ListNode next) { this.val = val; this.next = next; }
    }

    public static void main(String[] args) {
        ListNode head = new ListNode();
        ListNode p = head;
        p.next = new ListNode(1);
        p = p.next;
        p.next = new ListNode(1);
        p = p.next;
        p.next = new ListNode(2);

//        System.out.println("================");
//        while (head.next != null){
//            System.out.print(head.next.val);
//            if (head.next.next != null){
//                head = head.next;
//                System.out.print("->");
//            }else {
//                break;
//            }
//        }

        ListNode deleteDuplicates = deleteDuplicates2(head);

        System.out.println("================");
        while (deleteDuplicates.next != null){
            System.out.print(deleteDuplicates.next.val);
            if (deleteDuplicates.next.next != null){
                deleteDuplicates = deleteDuplicates.next;
                System.out.print("->");
            }else {
                break;
            }
        }

    }

    private static ListNode deleteDuplicates(ListNode head){
        ListNode pre = head.next;
        ListNode delete = pre.next;
        while (delete != null){
            if (delete.val == pre.val){
                pre.next = delete.next;
//                System.out.println("删除元素:"+delete.val);
                delete = delete.next;
            }else {
                pre = pre.next;
                delete = delete.next;
            }
        }
        return head;
    }

    //递归处理
    private static ListNode deleteDuplicates2(ListNode head){
        if (head == null || head.next == null){
            return head;
        }
        head.next = deleteDuplicates2(head.next);
        return head.val == head.next.val ? head.next : head;
    }
}