19. Remove Nth Node From End of List

题目

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.

思路

本题和上一道题目同是链表操作的题目，但是明显比上一掉题目Swap Nodes in Pairs简单，但是小技巧：建立指针头都是不变滴~。具体思路如下：

遍历链表，获得结点个数num，根据n找到需要删除的结点，将要删除结点的前一个结点指向要删除的结点的后一个结点。

代码

/**
* 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) {
ListNode* Head = head;
ListNode* helper = new ListNode(0);
helper->next = head;
ListNode* res = helper;
int num=0;
while(head)
{
num++;
head = head->next;
}
int diff = num -n;
while(diff)
{
diff--;
helper = helper->next;
Head = Head->next;
}
helper->next = Head->next;
return res->next;
}
};