LeetCode 82. Remove Duplicates from Sorted List II

题目

原题在此
Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

Example 2:

Input: head = [1,1,1,2,3]
Output: [2,3]

Constraints:

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

解析

1. 与Remove Duplicates from Sorted List I不同之处在于,这次要将重复值一个不留完全移除.
2. 有head的值会重复的情况,所以new一个值与head不一样的dummy指向head;
3. 声明两个指针, p1指向dummy, p2指向head;
4. 当p2不为空时,
   1. 当p2有next且next的值和p2相等时,p2指向next;
   2. 此时有两种情况,p1和p2挨着,两个指针分别指向自己的next. 否则p2指向自己的next,p1指向p2.此时便跳过了重复的值

      代码

      /**
       * Definition for singly-linked list.
       * struct ListNode {
       *     int val;
       *     ListNode *next;
       *     ListNode() : val(0), next(nullptr) {}
       *     ListNode(int x) : val(x), next(nullptr) {}
       *     ListNode(int x, ListNode *next) : val(x), next(next) {}
       * };
       */
      class Solution {
      public:
          ListNode* deleteDuplicates(ListNode* head) {
              if(!head) return head;
              ListNode dummy(head->val + 1, head);
              ListNode* p1 = &dummy, *p2 = head;
              while(p2) {
                  while(p2 -> next && p2 -> val == p2 -> next -> val) {
                      p2 = p2 -> next;
                  } 
                  if(p1 -> next == p2) {
                      p1 = p2;
                      p2 = p2 -> next;
                  } else {
                      p2 = p2 -> next;
                      p1 -> next = p2;
                  }
              }
              return dummy.next;
          }
      };