82. Remove Duplicates from Sorted List II

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每日一题 2019 - 04 - 16

题目：

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

Example 1:

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

Example 2:

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

解答：

这个题需要去除链表中所有重复的元素（只要重复该元素全部清除），思路比较直观也比较简单：

  • 开一个 vector 用来存放不重复的元素
  • 每次遍历链表时候判断当前位置的元素是否与 vector 顶端的元素相同，如果相同就跳过，且最后弹出 vector 最尾部元素
  • 最后需要注意，如果 vector 中一个元素也不存在，那么直接将 head 置空返回

代码：

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/**
 * 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 *temp = head ;
        vector<int> now ;
        now.push_back(temp -> val) ;
        temp = temp -> next ;
        while( temp != NULL )
        {
            if(now.size() == 0)
            {
                now.push_back(temp->val);
                temp = temp -> next ;
                continue ;
            }
            if(now[now.size()-1] == temp-> val)
            {
                while(now[now.size()-1] == temp-> val && temp != NULL)
                {
                    temp = temp -> next ;
                    if(temp == NULL)
                    {
                        break ;
                    }
                }
                now.pop_back();
            }
            else if(now[now.size()-1] != temp -> val)
            {
                now.push_back(temp->val);
                temp = temp -> next ;
            }
        }
        temp = head ;
        ListNode *before = NULL ;
        for(int i = 0 ; i < now.size() ; i ++)
        {
            temp -> val = now[i] ;
            before = temp;
            temp = temp ->next ;
        }
        if(now.size() == 0)
        {
            head = NULL ;
            return head ;
        }
        before -> next = NULL; 
        return head ;
    }
};
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