leetcode 40: 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.

 

 

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode ** cur = &head;
        ListNode *end = head;
        
        while( end !=NULL & n-->0) {
            end = end->next;
        }
        if(n>0) return head;
        
        while( end !=NULL) {
            end = end->next;
            cur = &((*cur)->next);
        }
        
        *cur = (*cur)->next;
        
        if( cur==&head) return *cur;
        else return head;
    }
};

\\alternate without using pointer to pointer.

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode * p = head;
        ListNode * pre = head;
        
        while( n-->0 && p!=NULL) {
            p=p->next;
        }
        
        if( n>0) return head;
        
        while(p!=NULL && p->next != NULL) {
            p=p->next;
            pre = pre->next;
        }
        
        if(pre == head) return pre->next;
        
        pre->next = pre->next->next;
        
        return head;
    }
};