LeetCode //C - 82. Remove Duplicates from Sorted List II-Toy模板网

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82. Remove Duplicates from Sorted List II

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

LeetCode //C - 82. Remove Duplicates from Sorted List II,LeetCode,leetcode,c语言,算法

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

Example 2:

LeetCode //C - 82. Remove Duplicates from Sorted List II,LeetCode,leetcode,c语言,算法

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

Constraints:

The number of nodes in the list is in the range [0, 300].
-100 <= Node.val <= 100
The list is guaranteed to be sorted in ascending order.

From: LeetCode
Link: 82. Remove Duplicates from Sorted List II

Solution:

Ideas：

1. Initialize Pointers and Dummy Node:

Create a dummy node that acts as a placeholder for the new list. Initialize prev to this dummy node and curr to the head of the original list.

2. Traverse the Original List:

Go through each node in the list using the curr pointer.

3. Check for Duplicates:

If the curr node has a duplicate (i.e., curr->val is the same as curr->next->val), move curr to the last duplicate node. This is done using the while loop within the main while loop.

4. Add Unique Nodes to New List:

If the curr node is not a duplicate, add it to the new list. The prev pointer keeps track of the last node added to the new list.

5. Move to Next Node:

Move curr to the next node in the original list.

6. Terminate New List:

Make sure to set the next of the last node in the new list to NULL.

7. Return New List:文章来源地址https://www.toymoban.com/news/detail-693751.html

The new list starts after the dummy node. So, we return dummy->next as the new head of the list.

Code:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) {
    // Create a dummy node to serve as the head of the new list
    struct ListNode* dummy = (struct ListNode*)malloc(sizeof(struct ListNode));
    dummy->val = 0;
    dummy->next = NULL;
    
    // Initialize pointers
    struct ListNode* prev = dummy;
    struct ListNode* curr = head;
    
    while (curr != NULL) {
        // Check if the current node is a duplicate
        int duplicate = 0;
        while (curr->next != NULL && curr->val == curr->next->val) {
            duplicate = 1;
            curr = curr->next;
        }
        
        // If it's not a duplicate, add it to the new list
        if (duplicate == 0) {
            prev->next = curr;
            prev = curr;
        }
        
        // Move to the next node in the original list
        curr = curr->next;
    }
    
    // Terminate the new list
    prev->next = NULL;
    
    // Return the new list (skipping the dummy node)
    struct ListNode* newHead = dummy->next;
    free(dummy);
    return newHead;
}