LeetCode - Remove Duplicates from Sorted List

最新推荐文章于 2025-09-11 19:19:55 发布

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文章标签：

#leetcode #linked list

本文介绍了一种从已排序链表中移除所有重复元素的方法，使得每个元素只出现一次。通过遍历链表并比较当前节点与下一个节点的值来实现，如果两节点值相同，则跳过下一个节点；若不同，则继续遍历。

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example, given 1->1->2, return 1->2; given 1->1->2->3->3, return 1->2->3.

http://oj.leetcode.com/problems/remove-duplicates-from-sorted-list/

Solution:

Pretty straight forward. Start from head and check if next node -> val is equal to current node -> val.

If we need to remove the next node, change the current -> next to current -> next -> next. Be careful about some edge cases.

https://github.com/starcroce/leetcode/blob/master/remove_duplicates_from_sorted_list.cpp

// 76 ms for 164 test case
// Given a sorted linked list, delete all duplicates such that each element appear only once.
/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(head == NULL) {
            return NULL;
        }
        ListNode *node = head;
        ListNode *next_node = head->next;
        while(next_node != NULL) {
            if(node->val == next_node->val) {
                ListNode *tmp = next_node;
                next_node = next_node->next;
                node->next = next_node;
                continue;
            }
            node = node->next;
            next_node = next_node->next;
        }
        return head;
    }
};