LeetCode:Remove Nth Node From End of List

problem:

Given a linked list, remove the n^th 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:解决方式采用双指针，前后指针相差n

/**
 * 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) {
    //题意是要删除从最后一个节点往前数的第n个节点 考虑到只有一个节点时没有办法删除 加一个dummy
    
    ListNode *dummy=new ListNode(-1);
    dummy->next=head;
    ListNode *pre=dummy;
    ListNode *last=dummy;
    
    //找到last的初始位置
    while(n--)
    {
        last=last->next;   
    }
    
    while(last->next!=NULL)
    {
        pre=pre->next;
        last=last->next;
    }
    //删除节点 
    ListNode *object=pre->next;
    pre->next=pre->next->next;
    delete object;
    
    return dummy->next;
    
    }
};