【LeetCode】82. 删除排序链表中的重复元素 II

给定一个排序链表,删除所有含有重复数字的节点,只保留原始链表中 没有重复出现 的数字。

基本思路是第一层循环,两个指针分别指向当前元素和它的前继,目的是如果有重复,需要删除当前元素,然后直到当前元素为空

第二层循环,两个指针分别指向当前元素的下一个元素和它的前继,目的是如果有重复,需要删除当前元素之后的所有元素,然后直到当前比较元素为空

ListNode* Solution::deleteDuplicates(ListNode* head)
{
    ListNode *pnewhead = new ListNode(0);
    ListNode *pcurrentlast = pnewhead;
    ListNode *pcurrent = head;
    unsigned int DuplicateFlag = 0;
    
    if((head == NULL)||(head->next == NULL))
    {
        return head;
    }
    pcurrentlast->next = head;
    while(pcurrent != NULL)
    {
        ListNode *pnext = pcurrent->next;
        ListNode *pnextlast = pcurrent;
        while(pnext != NULL)
        {
            if(pnext->val == pcurrent->val)
            {
                DuplicateFlag = 1;
                pnextlast->next = pnext->next;
                pnext = pnext->next;
            }
            else
            {
                pnext = pnext->next;
                pnextlast = pnextlast->next;
            }
        }
        /*no data duplicate*/
        if(0 == DuplicateFlag)
        {
            pcurrent = pcurrent->next;
            pcurrentlast = pcurrentlast->next;
        }
        else
        {
            /*data duplicate,delete current node*/
            pcurrentlast->next = pcurrent->next;
            pcurrent = pcurrent->next;
            DuplicateFlag = 0;
        }
    }
    return pnewhead->next;
}

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转载自blog.csdn.net/syc233588377/article/details/85245185