【LeetCode】19

Given a linked list, remove the nth node from the end of list and return its head.

For example,

   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.

Note:
Given n will always be valid.
Try to do this in one pass.

Solution 1: 重点在于one pass，以空间换取时间，使用vector<ListNode*>记录list的节点，每个指针指向一个链表节点，遍历一遍list算出size，通过下标计算（size-n）即可定位到需要删除的节点

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     ListNode* removeNthFromEnd(ListNode* head, int n) {　　//runtime:12ms
12         vector<ListNode*> vec;
13         ListNode *temp=head;
14         int size=0;
15         while(temp){
16             vec.push_back(temp);
17             temp=temp->next;
18             size++;
19         }
20         vec.push_back(NULL);//必不可少，否则如果删除的是倒数第一个元素的话下面的vec[i-1]->next=vec[i+1]中i+1溢出
21         int i=size-n;
22         if(i==0)//delete head
23             return head->next;
24         else{
25             vec[i-1]->next=vec[i+1];
26             return head;
27         }
28     }
29 };

Solution 2: 不需要额外的空间，当正数第n个移到链表尾时，head移到倒数第n个

 1 class Solution {
 2 public:
 3     ListNode* removeNthFromEnd(ListNode* head, int n) {　　　　//runtime:4ms
 4         ListNode *temp=head;
 5         while(--n)temp=temp->next;
 6         ListNode *del=head,*predel=NULL;
 7         while(temp->next){
 8             temp=temp->next;
 9             predel=del;
10             del=del->next;
11         }
12         if(predel==NULL)return head->next;
13         else{
14             predel->next=del->next;
15             return head;
16         }
17         
18     }
19 };