Leetcode Remove Nth Node from End of List

1 /**2 * id:193 * Name:remove Nth Node from End of List4 * Data structure:linked List5 * Time Complexity:6 * Space Complexity:7 * Tag:linklist8 * Difficult:easy9  Ten * Problem: One * Given A linked list, remove the nth node from the end of the list and return its head. A * Given linked list:1->2->3->4->5, and n = 2. - * After removing the second node from the end, the linked list becomes 1->2->3->5. - * Try to do this in one pass the  - * Solution: - * 1 -- - * Just run one pass to get the length of the list and then caculate the position from the beginning. + * then delete that item. - * 2 -- + * Use one vector to store the every Node and then we can use the "index in the" array to operate the link A * List directly. at  - * What to learn: - * When to consider give some cur->next value, firstly you should consider the original value maybe NULL. -  - * Definition for singly-linked list. - * struct ListNode { in * int val; - * ListNode *next; to * ListNode (int x): Val (x), Next (NULL) {} +  * }; -  **/ the  *#include <iostream> $#include <vector>Panax Notoginseng using namespacestd; -  the  + structListNode { A     intVal; theListNode *Next; +ListNode (intx): Val (x), Next (NULL) {} - }; $  $ voidPrint (ListNode *l) - { -ListNode *head =l; the      while(head!=NULL) -     {Wuyicout<< head->val<<Endl; theHead = head->Next; -     } Wu } -  About /* $ class Solution { - Public : - ListNode *removenthfromend (listnode *head, int n) { -           A ListNode *fakenode = new ListNode (0); + fakenode->next = head; the int length = 0; - ListNode * p = fakenode; $ While (p) the         { the length++; the p = p->next; the         } - length--; in delete p; the int cnt = length-n; the ListNode *cur = Fakenode; About While (CNT) the         { the cur=cur->next; the CNT--; +         } - if (cur->next) the Cur->next = cur->next->next;Bayi return fakenode->next; the     } the }; - */ -  the classSolution { the  Public: theListNode *removenthfromend (ListNode *head,intN) { the         if(Head = =NULL) -             returnNULL; the           theStd::vector<listnode *>v; theListNode *p =head;94          while(P) the         { the V.push_back (p); thep = p->Next;98         }  About         //cout<<v.size (); -         intPosition = V.size ()-N;101         if(Position = =0)102             returnV[position]->Next;103         Else if(Position = = V.size ()-1)104v[position-1]->next =NULL; the         Else106v[position-1]->next = v[position]->Next;107         returnv[0];108 109     } the };111  thelistnode* Create_list (int* Array,intN)113 { the     if(n==0) the         returnNULL; theListNode *head,*Current ;117Head = current =NewListNode (array[0]);118      for(intI=1; i<n;i++) 119     { -ListNode *node =NewListNode (Array[i]);121Current-and Next =node;122Current = currentNext;123     }124     returnhead; the }126 127  - intMain ()129 { the     //vector <int> a = {1,3,5};131 solution ans; the     //int a[3] = {1,3,5};133     intb[1] = {1};134ListNode * first = Create_list (b,1);135     //ListNode * second = create_list (a,3);136ListNode * Pure = Ans.removenthfromend (First,1);137     if(Pure = =NULL)138printf"NULL");139     Else $ print (pure);141     return 0;142}

Requirements: If you want to use a traversal, then open a new vector, which will save each node's address, easy to operate.

Leetcode Remove Nth Node from End of List