Question

3063. Linked List Frequency

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Description

Given the head of a linked list containing k distinct elements, return _the head to a linked list of length _k_ containing the frequency of each distinct element in the given linked list in any order._

 

Example 1:

Input: head = [1,1,2,1,2,3]

Output: [3,2,1]

Explanation: There are 3 distinct elements in the list. The frequency of 1 is 3, the frequency of 2 is 2 and the frequency of 3 is 1. Hence, we return 3 -> 2 -> 1.

Note that 1 -> 2 -> 3, 1 -> 3 -> 2, 2 -> 1 -> 3, 2 -> 3 -> 1, and 3 -> 1 -> 2 are also valid answers.

Example 2:

Input: head = [1,1,2,2,2]

Output: [2,3]

Explanation: There are 2 distinct elements in the list. The frequency of 1 is 2 and the frequency of 2 is 3. Hence, we return 2 -> 3.

Example 3:

Input: head = [6,5,4,3,2,1]

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 <= 105

Solutions

Solution 1: Hash Table

We use a hash table cnt to record the occurrence times of each element value in the linked list, then traverse the values of the hash table to construct a new linked list.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the linked list.

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def frequenciesOfElements(self, head: Optional[ListNode]) -> Optional[ListNode]:
    cnt = Counter()
    while head:
      cnt[head.val] += 1
      head = head.next
    dummy = ListNode()
    for val in cnt.values():
      dummy.next = ListNode(val, dummy.next)
    return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *   int val;
 *   ListNode next;
 *   ListNode() {}
 *   ListNode(int val) { this.val = val; }
 *   ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
  public ListNode frequenciesOfElements(ListNode head) {
    Map<Integer, Integer> cnt = new HashMap<>();
    for (; head != null; head = head.next) {
      cnt.merge(head.val, 1, Integer::sum);
    }
    ListNode dummy = new ListNode();
    for (int val : cnt.values()) {
      dummy.next = new ListNode(val, dummy.next);
    }
    return dummy.next;
  }
}

/**
 * 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 Solution {
public:
  ListNode* frequenciesOfElements(ListNode* head) {
    unordered_map<int, int> cnt;
    for (; head; head = head->next) {
      cnt[head->val]++;
    }
    ListNode* dummy = new ListNode();
    for (auto& [_, val] : cnt) {
      dummy->next = new ListNode(val, dummy->next);
    }
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func frequenciesOfElements(head *ListNode) *ListNode {
  cnt := map[int]int{}
  for ; head != nil; head = head.Next {
    cnt[head.Val]++
  }
  dummy := &ListNode{}
  for _, val := range cnt {
    dummy.Next = &ListNode{val, dummy.Next}
  }
  return dummy.Next
}

/**
 * Definition for singly-linked list.
 * class ListNode {
 *   val: number
 *   next: ListNode | null
 *   constructor(val?: number, next?: ListNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 *   }
 * }
 */

function frequenciesOfElements(head: ListNode | null): ListNode | null {
  const cnt: Map<number, number> = new Map();
  for (; head; head = head.next) {
    cnt.set(head.val, (cnt.get(head.val) || 0) + 1);
  }
  const dummy = new ListNode();
  for (const val of cnt.values()) {
    dummy.next = new ListNode(val, dummy.next);
  }
  return dummy.next;
}