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  • No.206 Reverse Linked List

    No.206 Reverse Linked List

    Reverse a singly linked list.

    单链表带头结点【leetcode判错】:

     1 #include "stdafx.h"
     2 
     3 struct ListNode
     4 {
     5     int val;
     6     ListNode *next;
     7     ListNode(int x):val(x),next(NULL){}
     8 };
     9 class Solution
    10 {
    11 public:
    12     ListNode* reverseList(ListNode* head)
    13     {//单链表反转
    14      //带头结点的单链表:head里面存放链表长度(或其他信息),head->next指向第一个实际节点;
    15         if(!(head) || !(head->next))//如果head为空,或者头结点指向空节点(链表长度为0)
    16             return head;
    17 
    18         ListNode *current = head->next;
    19         ListNode *back = NULL;//逆转之后,头结点变为尾节点,其next为Null  
    20         ListNode *front;
    21         //current 记录当前位置,back记录上一个位置,为current->next的值;front记录下一个位置,反转后current不等于current->next  
    22         while(current)//败给你了,不是!current
    23
    { 24 front = current->next;//反转后不能再用current->next,所以先记录下这个节点 25 current->next = back; 26 back = current; 27 current = front; 28 } 29 head->next = back; //current已经为空,所以back为尾节点。head->next指向它。 30 return head; 31 } 32 };

    不带头结点的单链表反转

    Input:[1,2]

    Output:[]

    Expected:[2,1]

     1 class Solution
     2 {
     3 public:
     4     ListNode* reverseList(ListNode* head)
     5     {//单链表反转
     6      //不带头结点的单链表
     7         if(!(head) || !(head->next))
     8             return head;
     9 
    10 //改:    ListNode *current = head->next;
    11         ListNode *current = head;
    12         ListNode *back = NULL;
    13         ListNode *front;
    14 
    15         while(current)//败给你了,不是!current
    16         {
    17             front = current->next;
    18             current->next = back;
    19             back = current;
    20             current = front;
    21         }
    22 //改:    head->next = back;
    23         head = back;
    24         return head;
    25     }
    26 };
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  • 原文地址:https://www.cnblogs.com/dreamrun/p/4528008.html
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