82. Remove Duplicates from Sorted List II && i

题目

83. Remove Duplicates from Sorted List

 Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3. 

解析

• [LeetCode] Remove Duplicates from Sorted List 移除有序链表中的重复项

class Solution_83 {
public:
	ListNode *deleteDuplicates(ListNode *head) {

		if (!head||!head->next)
		{
			return head;
		}

		ListNode* cur = head;
		ListNode*pre = NULL;
		while (cur&&cur->next)
		{
			pre = cur;
			cur = cur->next;
			ListNode* temp = pre; //记录每次重复点的开始位置
		    while(cur&&pre->val==cur->val)
			{
				pre = cur;
				cur=cur->next;
			}
			temp->next = cur; //跳过重复位置
		}
		return head;
	}
};

82. 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. 

解析

• 由于链表开头可能会有重复项，被删掉的话头指针会改变，而最终却还需要返回链表的头指针。所以需要定义一个新的节点，然后链上原链表，然后定义一个前驱指针和一个现指针，每当前驱指针指向新建的节点，现指针从下一个位置开始往下遍历，遇到相同的则继续往下，直到遇到不同项时，把前驱指针的next指向下面那个不同的元素。如果现指针遍历的第一个元素就不相同，则把前驱指针向下移一位。

//参考容易理解一些
ListNode *deleteDuplicates(ListNode *head) {
        if (!head || !head->next) return head;
        
        ListNode *start = new ListNode(0);
        start->next = head;
        ListNode *pre = start;
        while (pre->next) {
            ListNode *cur = pre->next;
            while (cur->next && cur->next->val == cur->val) cur = cur->next;
            if (cur != pre->next) pre->next = cur->next;
            else pre = pre->next;
        }
        return start->next;
    }


// 82. Remove Duplicates from Sorted List II
class Solution_82 {
public:
	ListNode* deleteDuplicates(ListNode* head) {

		if (!head||!head->next)
		{
			return head;
		}

		ListNode*newHead = new ListNode(0);
		newHead->next = head;

		ListNode* pre = newHead;
		ListNode* cur = head;
		
		while (cur&&cur->next)
		{
			ListNode* next = cur->next;
		
			if(next->val!=cur->val)
			{
				if (pre->next==cur) //pre->next当前元素开始，cur当前元素结束，cur->next另外不同的元素
				{
					pre = cur;
				}
				else
				{
					pre->next = cur->next;
				}
			}
			cur = cur->next;
		}
		if (pre->next!=cur) //这里是地址比较，若没有重复元素，则地址相同的
		{
			pre->next = cur->next;
		}
		return newHead->next;
	}
};

题目来源

• 82. Remove Duplicates from Sorted List II && i