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
p1
points to the current node. p
points to the
node before p1
so that we can ditch p1
if
needed.
The list is sorted so we only need dupVal
to keep the latest
duplicate value.
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var deleteDuplicates = function(head) {
if (!head) { return null }
const prehead = { next: head }
let p = prehead
let dupVal = NaN
for (let p1 = head; p1; p1 = p1.next) {
if (p1.val === dupVal) {
p.next = p1.next
} else if (p1.next && p1.val === p1.next.val) {
p.next = p1.next
dupVal = p1.val
} else {
p = p1
}
}
return prehead.next
};
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