19. Remove Nth Node From End of List

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?

又是照着答案都能语法出错的题。。。说明自己的指针都不知道干啥呢~

 

/**
 * 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) {
        ListNode* q = head;
        ListNode* p = head;
        if(head==NULL||head->next==NULL){
            return NULL;
        }
        for(int i=0;i<n;i++){
            q=q->next;
        }if(q==NULL){
            return head->next;
        }
        q = q->next;
        while(q!=NULL){
            q=q->next;
            p=p->next;
        }
        p->next = p->next->next;
        return head;
    }
};

这是榜首的那个人的代码

/**
 * 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) {
        ListNode *p = head,*q;
        q = p;
        int i=0,j=0;
        while(q != NULL) {
            q = q->next;
            ++i;
        }
        //cout << i << endl;
        while(j<=i-n-1){
            ++j;
            if(j>1) p = p->next;
        }
        cout << i<<p->val<< j<<endl;
        if(j == 0) {
            head = head->next;
            return head;
        } else {
        // p->next = NULL
           // p->next = p->next->next;
                q = p->next;
                p->next = q->next;
                //free(q);
                return (head);
        } 
    }
};

 cout << i<<p->val<< j<<endl;

不知道这句话是什么黑科技让时间少了一半