【LeetCode】19 - Remove Nth Node From End of List

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

/**
* Definition for singly-linked list.
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {　　//runtime:12ms
vector<ListNode*> vec;
ListNode *temp=head;
int size=0;
while(temp){
vec.push_back(temp);
temp=temp->next;
size++;
}
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
int i=size-n;
if(i==0)//delete head
return head->next;
else{
vec[i-1]->next=vec[i+1];
return head;
}
}
};

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

class Solution {
public:
ListNode* removeNthFromEnd(ListNode* head, int n) {　　　　//runtime:4ms
ListNode *temp=head;
while(--n)temp=temp->next;
ListNode *del=head,*predel=NULL;
while(temp->next){
temp=temp->next;
predel=del;
del=del->next;
}
if(predel==NULL)return head->next;
else{
predel->next=del->next;
return head;
}

}
};