Leetcode: Remove Duplicates from Sorted List II 解题报告

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.

SOLUTION 1:

使用一个del标记来删除最后一个重复的字元。

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode deleteDuplicates(ListNode head) {
         if (head == null) {
             return null;
         }

         // record the head.
         ListNode dummy = new ListNode(0);
         dummy.next = head;

         ListNode cur = dummy;

         // to delete the last node in the list of duplications.
         boolean del = false;

         while (cur != null) {
             if (cur.next != null
                 && cur.next.next != null
                 && cur.next.val == cur.next.next.val) {
                 cur.next = cur.next.next;
                 del = true;
             } else {
                 if (del) {
                     cur.next = cur.next.next;
                     del = false;
                 } else {
                     cur = cur.next;
                 }
             }
         }

         return dummy.next;
     }
 }

SOLUTION 2:

使用一个pre, 一个cur来扫描，遇到重复的时候，使用for循环用cur跳过所有重复的元素。

 /**
  * Definition for singly-linked list.
  * public class ListNode {
  *     int val;
  *     ListNode next;
  *     ListNode(int x) {
  *         val = x;
  *         next = null;
  *     }
  * }
  */
 public class Solution {
     public ListNode deleteDuplicates(ListNode head) {
         if (head == null) {
             return null;
         }

         ListNode dummy = new ListNode(0);
         dummy.next = head;

         ListNode pre = dummy;
         ListNode cur = pre.next;

         while (cur != null && cur.next != null) {
             if (cur.val == cur.next.val) {
                 while (cur != null && cur.val == pre.next.val) {
                     cur = cur.next;
                 }

                 // delete all the duplication.
                 pre.next = cur;
             } else {
                 cur = cur.next;
                 pre = pre.next;
             }
         }

         return dummy.next;
     }
 }