For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.
Idea: Use a pointer "pre" to traverse the whole list. Whenever pre.next.val==pre.val, skip the next node by pre.next=pre.next.next; otherwise push pre forward one step pre=pre.next.
Time: O(n) Space: O(1)
Code:
public class Solution {
public ListNode deleteDuplicates(ListNode head) {
if(head==null||head.next==null)
return head;
ListNode pre=head;
while(pre.next!=null)
{
if(pre.next.val==pre.val)
pre.next=pre.next.next;
else
pre=pre.next;
}
return head;
}
}
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