Question

1836. 从未排序的链表中移除重复元素

English Version

题目描述

给定一个链表的第一个节点 head ，找到链表中所有出现多于一次的元素，并删除这些元素所在的节点。

返回删除后的链表。

 

示例 1:

输入: head = [1,2,3,2]
输出: [1,3]
解释: 2 在链表中出现了两次，所以所有的 2 都需要被删除。删除了所有的 2 之后，我们还剩下 [1,3] 。

示例 2:

输入: head = [2,1,1,2]
输出: []
解释: 2 和 1 都出现了两次。所有元素都需要被删除。

示例 3:

输入: head = [3,2,2,1,3,2,4]
输出: [1,4]
解释: 3 出现了两次，且 2 出现了三次。移除了所有的 3 和 2 后，我们还剩下 [1,4] 。

 

提示：

• 链表中节点个数的范围是 [1, 105]
• 1 <= Node.val <= 105

解法

方法一：哈希表

我们可以用哈希表 $cnt$ 统计链表中每个元素出现的次数，然后遍历链表，删除出现次数大于 1 的元素。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为链表的长度。

# Definition for singly-linked list.
# class ListNode:
#   def __init__(self, val=0, next=None):
#     self.val = val
#     self.next = next
class Solution:
  def deleteDuplicatesUnsorted(self, head: ListNode) -> ListNode:
    cnt = Counter()
    cur = head
    while cur:
      cnt[cur.val] += 1
      cur = cur.next
    dummy = ListNode(0, head)
    pre, cur = dummy, head
    while cur:
      if cnt[cur.val] > 1:
        pre.next = cur.next
      else:
        pre = cur
      cur = cur.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 deleteDuplicatesUnsorted(ListNode head) {
    Map<Integer, Integer> cnt = new HashMap<>();
    for (ListNode cur = head; cur != null; cur = cur.next) {
      cnt.put(cur.val, cnt.getOrDefault(cur.val, 0) + 1);
    }
    ListNode dummy = new ListNode(0, head);
    for (ListNode pre = dummy, cur = head; cur != null; cur = cur.next) {
      if (cnt.get(cur.val) > 1) {
        pre.next = cur.next;
      } else {
        pre = cur;
      }
    }
    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* deleteDuplicatesUnsorted(ListNode* head) {
    unordered_map<int, int> cnt;
    for (ListNode* cur = head; cur; cur = cur->next) {
      cnt[cur->val]++;
    }
    ListNode* dummy = new ListNode(0, head);
    for (ListNode *pre = dummy, *cur = head; cur; cur = cur->next) {
      if (cnt[cur->val] > 1) {
        pre->next = cur->next;
      } else {
        pre = cur;
      }
    }
    return dummy->next;
  }
};

/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *   Val int
 *   Next *ListNode
 * }
 */
func deleteDuplicatesUnsorted(head *ListNode) *ListNode {
  cnt := map[int]int{}
  for cur := head; cur != nil; cur = cur.Next {
    cnt[cur.Val]++
  }
  dummy := &ListNode{0, head}
  for pre, cur := dummy, head; cur != nil; cur = cur.Next {
    if cnt[cur.Val] > 1 {
      pre.Next = cur.Next
    } else {
      pre = cur
    }
  }
  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 deleteDuplicatesUnsorted(head: ListNode | null): ListNode | null {
  const cnt: Map<number, number> = new Map();
  for (let cur = head; cur; cur = cur.next) {
    const x = cur.val;
    cnt.set(x, (cnt.get(x) ?? 0) + 1);
  }
  const dummy = new ListNode(0, head);
  for (let pre = dummy, cur = head; cur; cur = cur.next) {
    if (cnt.get(cur.val)! > 1) {
      pre.next = cur.next;
    } else {
      pre = cur;
    }
  }
  return dummy.next;
}