Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

Solution1：Iterative 依次比较 实时删除重复的

思路1_a：cur和prev相比较，若相同则delete cur；Note: 不应创建dummy从而从第一位开始，因为加入dummy的值可能影响重复，所以直接从第二位list开始。
思路1_b：cur和cur.next相比较
Time Complexity: O(N) Space Complexity: O(1)

Solution2：Iterative Two pointers "复写" (连接不重复的)

思路类似26题 http://www.jianshu.com/p/08a020a51fb3
Time Complexity: O(N) Space Complexity: O(1)

Solution3：Recursive 先序/后序 写法

思路3_a：递归。先序处理当前，再递归
思路3_b：递归。先递归，后序处理当前
Time Complexity: O(N) Space Complexity: O(N) 递归缓存

Solution1a Code：

class Solution1a {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null) return head;
        // if(head == null || head.next == null) return head;
        
        ListNode prev = head;
        ListNode cur = head.next;
        
        while(cur != null) {
            ListNode next_save = cur.next;   // for full delete
            if(cur.val == prev.val) {
                prev.next = cur.next;   
                cur.next = null;    // for full delete
            }
            else {
                prev = cur;
            }
            cur = next_save;
        }
        return head;
    }
}

Solution1b Code：

class Solution1b {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null) return head;
        
        ListNode cur = head;
        while(cur.next != null) {
            if(cur.val == cur.next.val) {
                ListNode save_next = cur.next; // for fully deletion
                cur.next = cur.next.next;
                save_next.next = null;  // for fully deletion
            }
            else {
                cur = cur.next;
            }
        }
        return head;
    }
}

Solution2 Code：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null || head.next == null) return head;
        
        ListNode new_cur = head;
        ListNode cur = head.next;
        while(cur != null) {
            if(cur.val != new_cur.val) {
                new_cur.next = cur;
                new_cur = cur;
            }
            // ListNode save_next = cur.next;   [for fully deletion]
            // cur.next = null;
            // cur = save_next;
            cur = cur.next;
        }
        new_cur.next = null;   //cut
        
        return head;
    } 
}

Solution3a Code：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null) return head;
        
        if(head.next.val == head.val) { // if duplicated
            head.next = head.next.next;
            head = deleteDuplicates(head); //restart
        }
        else {  // if not duplicated
            head.next = deleteDuplicates(head.next); // move next
        }
        return head;
    } 
}

Solution3b Code：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null) return head;
        head.next = deleteDuplicates(head.next);
        return head.val == head.next.val ? head.next : head;
    } 
}