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

    The 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.

    My analysis:

    When sovling linkedlist problems, two boundary cases needed to be very carefully treated, otherwise it may result in head lossing or get null pointer exception. 
    1. the head node might be changed. 
    Thus we may lose the information for new result linked list. One classic way is to use a dummy node to always pointer the the head the list. 
    In this problem, when duplicates appear, we would keep one copy. Thus the head node could never be changed, we won't need to set a dummy head for this case.
    
    2. If we have a scaner over the linkedlist, or several scanner(which we only need to consider the faster scanner). 
    The usually checking condition when traversaling the list is "while (faster != null)", In this case, we could only guarantee at the moment of entering the while loop, the faster is not null pointer. However, we may move the faster in the while loop:
    like:
    while (faster != null && ptr1.val == faster.val) {
        faster = faster.next;
    }
    the faster may reach the end of the list, thus faster is null.
    if we have a statement: "fast = faster.next" without any checking, it would result in a null pointer exception. 
    
    ???if we add a checking over here, it seems to distort the logic !
    In fact, the boundary case is the ending procedure over the list, after it, the invariant is over, we would not need to care about that!!!

    My first solution:

           if (head == null)
                return null;
            ListNode ptr1 = head;
            ListNode ptr2 = head.next;
            
            while (ptr2 != null) {
                while (ptr2 != null && ptr1.val == ptr2.val) {
                    ptr2 = ptr2.next;
                }
                ptr1.next = ptr2;
                ptr1 = ptr2;
                if (ptr2 != null) //may directly reach the end
                    ptr2 = ptr2.next;
            }
            return head;

    My elegant solution:

    public class Solution {
        public ListNode deleteDuplicates(ListNode head) {
            if (head == null)
                return null;
            ListNode dummy = new ListNode(0);
            ListNode pre = dummy;
            ListNode cur = head;
            while (cur != null) {
                if (pre == dummy || pre.val != cur.val) {
                    pre.next = cur;
                    pre = cur;
                }
                cur = cur.next;
            }
            pre.next = null; //remember the last step!!!
            return dummy.next;
        }
    }
  • 相关阅读:
    ERP类系统设计学习
    人工智能关键词
    系统性能
    连接不同服务器不同数据库
    socket一个例子
    SQLite
    asp.net 页面缓存、数据缓存
    原生js
    Android 网络调试 adb tcpip 开启方法
    C语言中string char int类型转换
  • 原文地址:https://www.cnblogs.com/airwindow/p/4255282.html
Copyright © 2011-2022 走看看