[LeetCode] 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.

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.

增加一个头结点，解决有可能需要删除原head问题，就不需要单独处理了。

/**
 * 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) {
        if(head==NULL || n==0)
            return head;
        ListNode *h = new ListNode(0);
        h->next = head;
        
        ListNode *p1 = h,*p2 = h;
        int num = 0;
        while(num!=n){
            p2 = p2->next;
            num++;
        }
        
        while(p2->next != NULL){
            p1 = p1->next;
            p2 = p2->next;
        }
        ListNode *p = p1->next->next;
        p1->next = p;
        return h->next;
        
    }