LeetCode_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.

 

/**
 * 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 *p, *q, *pre;
        p = q = head;
        int i = 1;
        while(q && i<n)
        {
            q = q->next;
            i++;
        }
        if(i < n) return head;
        
        while(q->next)
        {
            pre = p;
            p = p->next;
            q = q->next;
        }
        
        if(p == head)
            head = head->next;
          else
            pre->next = p->next;    
    
        delete p;
        return head;

    }
};

 

posted @ 2013-06-05 10:10  冰点猎手  阅读(138)  评论(0编辑  收藏  举报