【leetcode刷题笔记】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

.

题解：类似Remove Duplicates from Sorted List, 不过这次是如果有重复的数，那么把列表中所有的这个数都删除。

主要考察链表操作，建立一个新的链表头newNode，它的next指针指向head，head作为游标，初始指向newNode，然后当发现重复的时候，就可以利用一个while循环把重复的元素全部删除了，因为head总是指向要删除的节点前面一个节点。

代码如下：

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) {
*         val = x;
*         next = null;
*     }
* }
*/
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head == null || head.next == null)
return head;

ListNode newNode = new ListNode(0);
newNode.next = head;
head = newNode;

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

return newNode.next;
}
}