83. Remove Duplicates from Sorted List

原创于 2019-02-27 21:34:48 发布 · 169 阅读

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本文介绍了一种在排序链表中删除所有重复元素的方法，确保每个元素只出现一次。通过遍历链表并比较相邻节点的值，若发现重复则跳过重复节点，最终返回处理后的链表头。该方法时间复杂度为O(n)，空间复杂度为O(1)。

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题目描述

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

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

方法思路

Straight-Forward Approach
Complexity Analysis

Time complexity : O(n). 
Because each node in the list is checked exactly once to determine 
if it is a duplicate or not, the total run time is O(n), 
where nnn is the number of nodes in the list.

Space complexity : O(1). No additional space is used.

class Solution {
  public ListNode deleteDuplicates(ListNode head){
      ListNode handler = head;
      while(handler != null && handler.next != null){
          if(handler.val == handler.next.val)
              handler.next = handler.next.next;
              //跳过重复的结点
          else
              handler = handler.next;
      }
      return head;
  }
}