82 Remove Duplicates from Sorted List II

最新推荐文章于 2021-04-09 15:44:47 发布

原创最新推荐文章于 2021-04-09 15:44:47 发布 · 57 阅读

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本文提供了一种解决方案，用于从已排序的链表中移除所有重复出现的节点，仅保留原始列表中的不同数值。通过两个示例展示了该算法的工作原理及预期输出。

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1 题目

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

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

Input: 1->1->1->2->3
Output: 2->3

2 尝试解

class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* pre = dummy;
        ListNode* cur = pre->next;
        while(cur != NULL && cur->next != NULL){ //for example, 0->1->1->1->2->3
            if(cur->val != cur->next->val){  
                pre = pre->next;
                cur = cur->next;
            }
            else{  // cur is first 1,cur.next is second 1
                while(cur->next != NULL && cur->val == cur->next->val)  {//find the last 1             
                    cur = cur->next;
                }
                pre->next = cur->next; //0->2->3. 
                cur = pre->next;
            }
        }
        return dummy->next;
    }
};

3 标准解

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode **runner = &head;
        
        if(!head || !head->next)return head;
        
        while(*runner)
        {
            if((*runner)->next && (*runner)->next->val == (*runner)->val)
            {
                ListNode *temp = *runner;
                while(temp && (*runner)->val == temp->val)
                    temp = temp->next;
                
                *runner = temp;
            }
            else
                runner = &((*runner)->next);
        }
        
        return head;
    }
};