Remove Duplicates from Sorted List II

最新推荐文章于 2024-01-17 14:07:10 发布

原创最新推荐文章于 2024-01-17 14:07:10 发布 · 617 阅读

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本文详细阐述了如何解决给定排序链表中删除所有重复元素的问题，包括用例分析、边界条件判断、代码实现及优化策略。

首先我要在纸上，非常非常聪明且迅速且机灵，
给出几个用例，找出边界用例和特殊用例，确定特判条件；在编码前考虑到所有的条件
向面试官提问：问题规模，特殊用例
给出函数头
暴力解，简述，优化。
给出能够想到的最优价
伪代码，同时结合用例
真实代码

Remove Duplicates from Sorted List II

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

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

» Solve this problem

/*
Remove Duplicates from Sorted List II
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

*/

/*
Loop
Unsorted
NULL
1
1 1
1 1 2
1 2 2
1 2 2 3
*/

/**
 * 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 || head->next == NULL) return head;
    
    ListNode* p = head;
    ListNode* c = head->next;
    ListNode* n = NULL;
    
    
    bool dup_head = false;
    if(head->val == head->next->val)
    {
        dup_head = true;
        c = p;
    } 
    
    while(c!=NULL)
    {   
        
        bool flag = false;
        while(c->next != NULL && c->val == c->next->val)
        {
            flag = true;
            n = c->next;
            delete c;
            c = n;
        }    
        if(flag)
        {
            n = c->next;
            delete c;
            
            if(dup_head)
            {
                head = n;
                dup_head = false;
            }
            else
            {
                p->next = n;
            }
        }    
        c = n;
    
    }
    
    
    return head;
}
};