Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.
Example 1:
Input: head = [1,2,3,3,4,4,5] Output: [1,2,5]
Example 2:
Input: head = [1,1,1,2,3] Output: [2,3]
Constraints:
- The number of nodes in the list is in the range
[0, 300]. 100 <= Node.val <= 100- The list is guaranteed to be sorted in ascending order.
/**
* Definition for singly-linked list.
* function ListNode(val, next) {
* this.val = (val===undefined ? 0 : val)
* this.next = (next===undefined ? null : next)
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
const dummy = new ListNode(0);
dummy.next = head;
let cur = head;
let prev = dummy;
let duplicate = null;
while (cur && cur.next) {
if (cur.val === duplicate || cur.val === cur.next.val) {
duplicate = cur.val;
prev.next = cur.next;
} else {
prev = prev.next;
}
cur = cur.next;
}
if (cur && cur.val === duplicate) {
prev.next = null;
}
return dummy.next;
};
var deleteDuplicates = function(head) {
let pre = new ListNode(0);
pre.next = head;
let node = pre;
while (node.next && node.next.next) {
if (node.next.val === node.next.next.val) {
let val = node.next.val;
while (node.next && node.next.val == val) {
node.next = node.next.next;
}
} else {
node = node.next;
}
}
return pre.next;
};
Complexity Analysis
- Time complexity:
O(n) - Space complexity:
O(1)