19. 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.

题目链接

思路
只遍历一遍，这里就想到用两个指针，第一个先走n步，然后再一起遍历，第一个到达尾部的时候，第二个指向的刚好是倒数第n个，然后把它删掉即可。

代码（C）

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     struct ListNode *next; * }; */struct ListNode* removeNthFromEnd(struct ListNode* head, int n) {    struct ListNode *newhead = (struct ListNode *)malloc(sizeof(struct ListNode));    if (NULL == newhead)        return head;    newhead->next = head;    struct ListNode *fast = newhead;    struct ListNode *pre = NULL;    int i = 0;    while(i < n)    {        fast = fast->next;        i++;    }    pre = newhead;    while(fast->next != NULL)    {        pre = pre->next;        fast = fast->next;    }    pre->next = pre->next->next;    return newhead->next;}

这里定义一个newhead的原因是：当n等于链表长度时，需要删除头指针，这时候head要被删除。

代码（python）

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeNthFromEnd(self, head, n):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        newhead = ListNode(0)        newhead.next = head        fast = head        index = 0        while index < n:            fast = fast.next            index += 1        pre = newhead        while fast:            pre = pre.next            fast = fast.next        pre.next = pre.next.next        return newhead.next