Answer
See the explanation
Work Step by Step
a).To remove duplicates from an unsorted linked list, you can use a hash set to keep track of unique elements as you traverse the linked list. Here's a high-level algorithm:
1. Initialize an empty hash set to store unique elements.
2. Traverse the linked list:
- For each node, check if its value is already present in the hash set.
- If it is, remove the node from the linked list.
- If it's not, add the value to the hash set.
3. Continue traversing until the end of the linked list is reached.
By using a hash set, you can achieve a time complexity of O(n) since each lookup and insertion operation takes constant time on average. This algorithm ensures that each element in the linked list remains unique after duplicates are removed.
b).If the use of a temporary buffer is not permitted to solve the problem mentioned in part (a), an alternative approach can be used. One possible solution is to iterate through the linked list with two pointers, where one pointer moves at a slower pace (one step at a time) and the other pointer moves at a faster pace (two steps at a time).
By doing so, if there is a loop in the linked list, the faster pointer will eventually catch up to the slower pointer within the loop. This can be explained by considering the relative speeds of the two pointers. If there is a loop, the faster pointer will keep circling around the loop and eventually catch up to the slower pointer.
To determine if there is a loop, we can initialize two pointers, slow and fast, at the head of the linked list. Then, we can move the slow pointer one step at a time and the fast pointer two steps at a time. If the two pointers meet at any point, it indicates the presence of a loop in the linked list.
Once the loop is detected, we can find the starting point of the loop by resetting the slow pointer to the head of the linked list and moving both pointers at the same pace (one step at a time) until they meet again. The meeting point will be the starting point of the loop.
This approach works because the distance between the head of the linked list and the starting point of the loop is equal to the distance between the meeting point and the starting point of the loop. By moving both pointers at the same pace, they will meet at the starting point of the loop.
Therefore, if the use of a temporary buffer is not allowed, this approach using two pointers can be used to detect and find the starting point of a loop in a linked list.
