82. Remove Duplicates from Sorted List II
Tags: ‘Linked List’
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
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
Solution
Similar to 83, this time we need dummy node .
Also when no duplicate, prev shouldn’t increment
//Better. no need for after pointer
public ListNode deleteDuplicates(ListNode head) {
// need a dummy node this time for corner cases
if (head == null || head.next == null) return head;
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode prev = dummy, curr = head;
while (curr != null) {
while (curr.next != null && curr.next.val == curr.val) {
curr = curr.next;
}
if (prev.next == curr) { // no duplicate, increment both
prev = prev.next;
} else {
prev.next = curr.next;
}
curr = curr.next;
}
return dummy.next;
}
public ListNode deleteDuplicates(ListNode head) {
// need a dummy node this time for corner cases
if (head == null || head.next == null) return head;
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode prev = dummy, curr = head;
while (curr != null) {
ListNode after = curr.next;
while (after != null && after.val == curr.val) {
after = after.next;
}
if (curr.next == after) { // no duplicates, just increment both
prev = prev.next;
curr = curr.next;
} else { // delete duplicate, and set curr to after
prev.next = after;
curr = after;
}
}
return dummy.next;
}