zoukankan      html  css  js  c++  java
  • [Leetcode 5] 83 Remove Duplicates from Sorted List

    Problem:

    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.

    Analysis:

    It's similar to the remove duplicates from sorted array problem. Use pointers ptrA and ptrB. ptrA always points to the last position of non-duplicated list and ptrB go through the list. One important thing is that we need a third point always points to the previous position of ptrB to delete duplicates. The third pointer can ensure that those duplicated nodes have no reference point to them and thus can be GC cleaned as soon as possible.

    The time complexity is O(n) and the space complexity is O(n)

    Code:

    /**
     * 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) {
            // Start typing your Java solution below
            // DO NOT write main() function
            if (head == null) return null;
            
            ListNode cur=head, valid=head, prev=null;
            
            while (cur.next != null) {
                prev = cur;
                cur = cur.next;
                if (cur.val != valid.val) {
                    if (prev.val == valid.val) prev.next=null;
                    valid.next = cur;
                    valid = cur;
                } else {
                  prev.next = null;
                }
            }
            
            return head;
        }
    }

    Attention:

    To access Java field, use . not ->

  • 相关阅读:
    IDEA右键新建时没有Java Class选项
    捕获摄像头视频VC
    重叠IO与IOCP
    (八)内存管理与内存分配
    DebugView使用详解
    (六) 中断机制
    (五) proc文件系统
    bash 之备份文件
    bash 遍历目录文件
    (四) linux内核模块编程
  • 原文地址:https://www.cnblogs.com/freeneng/p/3003323.html
Copyright © 2011-2022 走看看