83.删除排序链表中的重复元素

对于重复的只取最后一个ListNode即可,通过递归依次去重

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

另外一种解法,取重复的第一个为current,然后如果是重复的就跳过。

//1->1->2->3->3
public ListNode deleteDuplicates(ListNode head) {
ListNode current = head;
while (current != null && current.next != null) {
if (current.next.val == current.val) {
//跳过
current.next = current.next.next;
} else {
current = current.next;
}
}
return head;
}