83. Remove Duplicates from Sorted List + 82. Remove Duplicates from Sorted List II

▶ 删除单链表中的重复元素。

▶ 83. 把重复元素删得只剩一个，如 1 → 1 → 2 → 3 → 3 → 3 → 4 → 5 → 5 变成 1 → 2 → 3 → 4 → 5。注意要点：第一个元素就可能重复，最后一个元素可能是重复，多个连续重复。

● 自己的代码，18 ms，记录了发生重复的第一个元素的位置，使其指向下一个不同的元素的位置。

 class Solution
 {
 public:
     ListNode* deleteDuplicates(ListNode* head)
     {
         if (head == nullptr || head->next == nullptr)
             return head;

         ListNode *output, *i, *j, *k;
         for (i = head, j = head->next; i != nullptr; i = j)
         {
             for (; j != nullptr && i->val == j->val; j = j->next);
             i->next = j;
         }
         return head;
     }
 };

● 大佬的代码，11 ms，逐格比较相邻两个结点，值相同则跨过后一结点。

 class Solution
 {
 public:
     ListNode* deleteDuplicates(ListNode* head)
     {
         for (ListNode* cur = head; cur && cur->next;)
         {
             if (cur->val == cur->next->val)
                 cur->next = cur->next->next;
             else
                 cur = cur->next;
         }
         return head;
     }
 };

▶ 82. 把发生重复的元素删除得一个不剩。如 1 → 1 → 2 → 3 → 3 → 3 → 4 → 5 → 5 变成 2 → 4 。

● 自己的非递归版代码，8 ms

 class Solution
 {
 public:
     ListNode* deleteDuplicates(ListNode* head)
     {
         if (head == nullptr || head->next == nullptr)
             return head;
         for (; head != nullptr && head->next != nullptr && head->val == head->next->val;)// 去掉开头的重复结点
         {
             for (; head->next != nullptr && head->val == head->next->val; head = head->next);
             head = head->next;
         }
         if (head == nullptr)
             return head;
         for (ListNode *prev = head, *cur = head->next; cur != nullptr && cur->next != nullptr; cur = prev->next)
         {
             if (cur->val == cur->next->val)
             {
                 for (; cur->next != nullptr && cur->val == cur->next->val; cur->next = cur->next->next);
                 prev->next = cur->next;
             }
             else
                 prev = prev->next;
         }
         return head;
     }
 };

● 大佬的递归版代码，9 ms

 class Solution
 {
 public:
     ListNode* deleteDuplicates(ListNode* head)
     {
         if (head == nullptr || head->next == nullptr)
             return head;

         int val = head->val;
         ListNode *p = head->next;
         if (p->val != val)
         {
             head->next = deleteDuplicates(p);
             return head;
         }
         else
         {
             for (; p && p->val == val; p = p->next);
             return deleteDuplicates(p);
         }
     }
 };