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 = None

class 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