Solution

Straight-Forward Approach [Accepted]

Algorithm

This is a simple problem that merely tests your ability to manipulate list node pointers. Because the input list is sorted, we can determine if a node is a duplicate by comparing its value to the node after it in the list. If it is a duplicate, we change the next pointer of the current node so that it skips the next node and points directly to the one after the next node.

Java

public ListNode deleteDuplicates(ListNode head) {\n    ListNode current = head;\n    while (current != null && current.next != null) {\n        if (current.next.val == current.val) {\n            current.next = current.next.next;\n        } else {\n            current = current.next;\n        }\n    }\n    return head;\n}\n

Complexity Analysis

Because each node in the list is checked exactly once to determine if it is a duplicate or not, the total run time is $$O(n)$$ , where $$n$$ is the number of nodes in the list.

Space complexity is $$O(1)$$ since no additional space is used.

Correctness

We can prove the correctness of this code by defining a loop invariant. A loop invariant is condition that is true before and after every iteration of the loop. In this case, a loop invariant that helps us prove correctness is this:

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All nodes in the list up to the pointer current do not contain duplicate elements.

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We can prove that this condition is indeed a loop invariant by induction. Before going into the loop, current points to the head of the list. Therefore, the part of the list up to current contains only the head. And so it can not contain any duplicate elements. Now suppose current is now pointing to some node in the list (but not the last element), and the part of the list up to current contains no duplicate elements. After another loop iteration, one of two things happen.

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1. \n

   current.next was a duplicate of current. In this case, the duplicate node at current.next is deleted, and current stays pointing to the same node as before. Therefore, the condition still holds; there are still no duplicates up to current.

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2. \n

   current.next was not a duplicate of current (and, because the list is sorted, current.next is also not a duplicate of any other element appearing before current). In this case, current moves forward one step to point to current.next. Therefore, the condition still holds; there are no duplicates up to current.

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At the last iteration of the loop, current must point to the last element, because afterwards, current.next = null. Therefore, after the loop ends, all elements up to the last element do not contain duplicates.

Analysis written by: @noran.