Problem
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 (Java)
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) {
return head;
}
ListNode dummy = new ListNode(0);
ListNode prev = dummy;
prev.next = head;
ListNode curr = head.next;
while (curr != null) {
boolean flagFoundDuplicate = false;
while (curr != null && prev.next.val == curr.val) {
flagFoundDuplicate = true;
curr = curr.next;
}
if (flagFoundDuplicate) {
prev.next = curr;
if (curr != null) {
curr = curr.next;
}
} else {
prev = prev.next;
prev.next = curr;
curr = curr.next;
}
}
return dummy.next;
}
}
Solution (Javascript)
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
var newHead = new ListNode(0);
var now = newHead;
var tmp = head;
var val = 0;
while (tmp) {
val = tmp.val;
if (tmp.next && tmp.next.val === val) {
tmp = tmp.next;
while (tmp && tmp.val === val) tmp = tmp.next;
} else {
now.next = tmp;
now = tmp;
tmp = tmp.next;
now.next = null;
}
}
return newHead.next;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).