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

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.

Follow up:

Could you do this in one pass?

class Solution {
public:  
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        //刪除倒數第n個轉化為，刪除正數第幾個
        //要考慮比較多的特殊情況，head==NULL head->next==NULL
        
 
//還有len-n==0的
        int len1=len(head);
        if(head==NULL ||head->next==NULL) return NULL;
        ListNode *p;
        p=head;
        if(len1-n==0)   head=head->next;
        else{
             for(int i=1;i<len1-n;i++){
                p=p->next;
             }
             p->next=p->next->next;
            }
            
 
return head;    
    }
    int len(ListNode *Head)
    {
        int count=0;
        while(Head!=NULL){
            Head=Head->next;
            count++;
        }
        return count;
    }
};

思路二：設定兩個指標，一個先走n步，然後兩個指標再一起走，等第一個指標到達連結串列最後一個結點，另一個指標指向結點的下一個結點，就是需要刪除的倒數第n個結點

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *p1,*p2;
        p1=head;
        p2=head;
        if(head==NULL || head->next==NULL) return NULL;
        while(n--){
            p1=p1->next;
        }
        if(p1==NULL) return head->next; //這種情況直接return,不要只是修改head,這樣後面繼續執行會出錯
        while(p1->next!=NULL){
            p1=p1->next;
            p2=p2->next;
        }
        p2->next=p2->next->next;
        return head;
    }
};