leetcode 83 Remove Duplicates from Sorted List-优快云博客

本文链接：https://blog.youkuaiyun.com/wangyaninglm/article/details/45876489

本文介绍如何通过遍历并比较链表节点值来删除排序链表中的重复元素，确保每个元素仅出现一次。

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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.
这里写图片描述

下面是我的解决方案，考虑测试用例：
1，1
1，1，1
1，2，2

/**
 * 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)
    {
        if(head==NULL)return head;

        ListNode* pre = head;
        ListNode* cur = head->next;

        while(cur!=NULL)
        {
            if(cur->val==pre->val)
            {
                pre->next = pre->next->next;
                cur = cur->next;
                if(cur==NULL)return head;
                //free(cur)没有free是不对的，可能引起内存泄漏;
            }
            else if(cur->val!=pre->val)
            {
                pre = pre->next;
                cur = cur->next;
            }

        }
        return head;

    }
};

c++：

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if (head == null) return head;

        ListNode cur = head;
        while(cur.next != null) {
            if (cur.val == cur.next.val) {
                cur.next = cur.next.next;
            }
            else cur = cur.next;
        }
        return head;
    }
}

c语言：

struct ListNode* deleteDuplicates(struct ListNode* head) 
{
    if (head) {
    struct ListNode *p = head;
    while (p->next) {
        if (p->val != p->next->val) {
            p = p->next;
        }
        else {
            struct ListNode *tmp = p->next;
            p->next = p->next->next;
            free(tmp);//这块在实际代码中，非常有必要
        }
    }
}

return head;


}

简洁的python解决方案：

class Solution:
    # @param head, a ListNode
    # @return a ListNode
    def deleteDuplicates(self, head):
        node = head
        while node:
            while node.next and node.next.val == node.val:
                node.next = node.next.next

            node = node.next

        return head