算法：删除链表倒数第n个数

LeetCode OJ Problem：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.

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        if(n == 0)
            return head;
        int num = n;
        ListNode *p, *pre_p, *q;
        p = pre_p = q = head;
        
        while(num > 1)
        {
            q = q->next;
            num--;
        }
        while(q->next != NULL)
        {
            pre_p = p;
            p = p->next;
            q = q->next;
        }
        if(p == head)//如果删除的是头结点
        {
            head = p->next;
            p->next = NULL;
        }
        else
        {
            pre_p->next = p->next;
            p->next = NULL;
        }
       
        delete p;
        return head;
        
    }
};

算法：删除链表倒数第n个数,布布扣,bubuko.com

算法：删除链表倒数第n个数

原文：http://blog.csdn.net/wan_hust/article/details/24330463