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Difficulty Level: Moderate
1. Description
Given the head of a sorted linked list head, remove all nodes with duplicate numbers in the original linked list, leaving only distinct numbers . Returns a sorted linked list .
2. Examples
Example 1
输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]
Example 2
输入：head = [1,1,1,2,3]
输出：[2,3]
Restrictions:
链表中节点数目在范围[0, 300]` within-100 <= Node.val <= 100- The topic data guarantees that the linked list has been arranged in ascending order
3. Answer
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class RemoveDuplicatesfromSortedListII {
func deleteDuplicates(_ head: ListNode?) -> ListNode? {
if head == nil || head!.next == nil {
return head
}
let dummy = ListNode(0)
dummy.next = head
var node = dummy
while node.next != nil && node.next!.next != nil {
if node.next!.val == node.next!.next!.val {
let val = node.next!.val
while node.next != nil && node.next!.val == val {
node.next = node.next!.next
}
} else {
node = node.next!
}
}
return dummy.next
}
}
- Main idea: iterate over the list, skipping duplicates by replacing next with next.next.
- Time complexity: O(n)
- Space Complexity: O(1)
Note: Swift provides "===" to compare two objects referring to the same reference
The warehouse of the algorithm solution: LeetCode-Swift
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