[LeetCode]Remove Duplicates from Sorted List II-优快云博客

本文链接：https://blog.youkuaiyun.com/sunbaigui/article/details/8981252

本文提供了一种使用C++实现的高效方法来移除链表中连续的重复元素。通过两个指针技巧，该方法能有效保留链表中唯一的元素，并且在处理各种边界情况时表现出良好的鲁棒性。

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struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};


class Solution {
//three pointer, the previous valid pointer, the first pointer of these same elements, current pointer
//always think twice about these edge cases 
public:
	ListNode *deleteDuplicates(ListNode *head) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		ListNode* preValid = NULL;
		ListNode* preSame = NULL;
		ListNode* cur = head;
		while (cur)
		{
			if(!preSame) preSame = cur;
			else if(cur->val != preSame->val && preSame->next == cur)//continuous different then preSame can be the preValid
			{
				if(!preValid)
				{
					head = preSame;
					preValid = preSame;
				}
				else
				{
					preValid->next = preSame;
					preValid = preSame;
				}
				preSame = cur;
			}
			else if (cur->val != preSame->val && preSame->next != cur)//not continuous different
				preSame = cur;
			cur = cur->next;
		}
		//4 special edge cases
		if(!preValid && preSame && preSame->next != NULL) return NULL;
		else if(!preValid && preSame && preSame->next == NULL) return preSame;
		else if(preValid && preSame->next != NULL) preValid->next = NULL;
		else if(preValid && preSame->next == NULL) preValid->next = preSame;
		
		return head;
	}
};

second time

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode dummy(-1);
        ListNode* savePtr = &dummy;
        ListNode* cur = head;
        ListNode* prev = &dummy;
        while(cur != NULL)
        {
            if((prev == &dummy || prev->val != cur->val) 
                && (cur->next == NULL || cur->next->val != cur->val))
            {
                savePtr->next = cur;
                savePtr = savePtr->next;
            }
            prev = cur;
            cur = cur->next;
        }
        savePtr->next = NULL;
        return dummy.next;
    }
};