82. Remove Duplicates from Sorted List II（删除给定有序链表的重复结点 II）

题目描述

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
给定已排序的链接列表，删除所有具有重复数字的节点，只留下原始列表中的不同数字。
Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

Input: 1->1->1->2->3
Output: 2->3

方法思路

只要是重复的，要将重复的元素一个不留的删除

Approach1:非递归的解决方案

class Solution {
    //Runtime: 0 ms, faster than 100.00% 
    //Memory Usage: 37.5 MB, less than 57.51%
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)
            return head;
        ListNode dummy = new ListNode(0);
        ListNode cur = head, pre = dummy;
        dummy.next = head;
        boolean flag = true;
        while(cur != null && cur.next != null){
            if(cur.val == cur.next.val){
                flag = false;
                cur = cur.next;
                continue;
            }else{
                if(!flag){
                    pre.next = cur.next;
                    flag = true;
                    cur = cur.next;
                }else{
                    pre = cur;
                    cur = cur.next;
                }
            }
        }
        if(flag == false)//针对以重复结点结束的链表
            pre.next = null;
        return dummy.next;
    }
}

上下两个代码是相同思路的不同实现，区别不大，下面的更简洁易懂一些

class Solution{
    //use two pointers, pre - track the node before the dup nodes, 
	//cur - to find the last node of dups.
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null) return null;
        ListNode dummy=new ListNode(0);
        dummy.next=head;
        ListNode pre=dummy;
        ListNode cur=head;
        while(cur!=null){
            //loop to find the last node of the dups.
            while(cur.next!=null&&cur.val==cur.next.val){
                cur=cur.next;
            }
            if(pre.next==cur){
                //no dup, move down both pointer.
                pre=pre.next;
            }
            else{
                //duplicates detected,remove the dups.
                pre.next=cur.next;
            }
            cur=cur.next;
        }
        return dummy.next;
    }
}

Approach2: recursive

class Solution{
    //Runtime: 0 ms, faster than 100.00%
    //if current node is not unique, return deleteDuplicates with head.next.
    //If current node is unique, link it to the result of next list made by recursive call.
    public ListNode deleteDuplicates(ListNode head) {
    if (head == null) return null;
    
    if (head.next != null && 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;
}
}