【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.

用双指针，使两个指针相隔n-1个数，当前面的指针指向末尾NULL的时候，p刚好指向要删除的那个节点。

/**
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (head==NULL) return NULL;
        ListNode *p = head;
        ListNode *q = head;
        ListNode *pPre = NULL;
        for(int i=0;i<n-1;i++){
            q=q->next;
        }
        while(q->next!=NULL){
            pPre = p;
            p=p->next;
            q=q->next;
        }
        if(pPre==NULL)
            head = p->next;
        else
            pPre->next=p->next;
        delete p;
        return head;
    }
};

　　

下面的做法是错误的，因为如果删不到第一个结点。所以应该有一个pPre来记录上一个结点的位置。

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if (head==NULL) return NULL;
        ListNode *p = head;
        ListNode *q = head;
        for(int i=0;i<n;i++){
            q=q->next;
        }
        while(q->next!=NULL){
            p=p->next;
            q=q->next;
        }
        ListNode *dp=p->next;
        p->next=p->next->next;
        delete dp;
        return head;
    }
};