82. Remove Duplicates from Sorted List II

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

Solution1：Iterative Two pointers "复写" (连接不重复的)

类似83题Solution2:

屏幕快照 2017-09-11 下午3.39.59.png

http://www.jianshu.com/p/ddeed81dc376

Time Complexity: O(N) Space Complexity: O(1)

Solution2：Recursive 先序 写法

思路：递归。先序处理当前，再递归
和第83题Solution3_a 思路相同 http://www.jianshu.com/p/ddeed81dc376
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution1 Code：

class Solution1 {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null) return null;
        ListNode dummy = new ListNode(0);
        dummy.next=head;
        
        ListNode prev = dummy;
        ListNode cur = head;
        while(cur != null) {
            //cur: skip the duplicates and find the next unique one
            while(cur.next != null && cur.val == cur.next.val){ 
                cur = cur.next;
                // ListNode save_next = cur.next;//   [for fully deletion]
                // cur.next = null;
                // cur = save_next;
            }
            
            if(cur != prev.next) { // there is duplicates, link cur.next      
                prev.next = cur.next;
            }
            else { // if no skipping, that is a unique one. Then continue
                prev = prev.next;
            }
            
            cur = cur.next;
        }
        return dummy.next;
    }
}

Solution2 Code：

class Solution2 {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null || head.next == null) return head;

        if (head.val == head.next.val) {
            while (head.next != null && head.val == head.next.val) {
                head = head.next;
            }
            return deleteDuplicates(head.next);
        } else {
            head.next = deleteDuplicates(head.next);
        }
        return head;
    }
}