,2,1] Output: [1,1,1,1,1,1] Explanation: There are `6` distinct elements in the list. The frequency of each of them is `1`. Hence, we return `1 -> 1 -> 1 -> 1 -> 1 -> 1`.

Constraints:

The number of nodes in the list is in the range [1, 105].
1 <= Node.val <= 10^5

Solution

Method 1 – Hash Map Frequency Count (1)

Intuition

We need to count the frequency of each distinct value in the linked list. Using a hash map allows us to efficiently count occurrences as we traverse the list, then we can build a new linked list from the frequency values.

Approach

Initialize an empty hash map to store frequencies.
Traverse the linked list, incrementing the count for

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