LeetCode的medium题集合(C++实现)十六

1 Remove Duplicates from Sorted Array II

Follow up for “Remove Duplicates”:What if duplicates are allowed at most twice?For example,Given sorted array nums = [1,1,1,2,2,3],

Your function should return length = 5, with the first five elements of nums being 1, 1, 2, 2 and 3. It doesn’t matter what you leave beyond the new length.

与删除重复的元素不同，这里只删除重复数超过2的部分。与删除重复的元素做法一样，只不过多用一个标记量flag记录重复的个数是否超过2.

int removeDuplicates(vector<int>& nums)
    {
        int n=nums.size();
        if(n==0)
        return 0;
        int length=0;
        bool flag=false;
        for(int i=1;i<n;i++)
        {
            if(nums[length]!=nums[i])
            {
                nums[length+1]=nums[i];
                length++;
                flag=false;
            }
            else
            {
                if(!flag)
                {
                    flag=true;
                    nums[length+1]=nums[i];
                    length++;
                }
            }
        }
        return length+1;
    }

2 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.

因为相同的元素全部被删除，则链表的第一个指针可能被删，因此必须要用到指向链表第一个节点的头指针。

 ListNode* deleteDuplicates(ListNode* head) {
         if(head==NULL)
        return NULL;
        ListNode* result=new ListNode(0);
        result->next=head;
        ListNode* pre=result;
        ListNode* cur=head;
        while(cur)
        {
            if(cur->next&&cur->val==cur->next->val)
            {
              int value=cur->val;
              while(cur->next&&value==cur->next->val)
             {
             //删除第一个元素以外的重复元素
               ListNode* mid=cur->next;
               cur->next=mid->next;
               delete mid;
             }
             ListNode* t=cur;
             cur=t->next;
             delete t;  //删除第一个重复元素
             pre->next=cur;
            }
            else
            {
               pre=cur;
               cur=cur->next;
            }
        }
        return result->next;
    }