Leetcode-19. Remove Nth Node From End of List

Leetcode-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.

本题可以利用快慢指针的方法来解决，让快指针先走n步，然后快指针和慢指针同时走，当快指针到达终点时，慢指针指向的就是我们要删除的节点了。
删除的节点的方式可以用多种表示，如若想删除1和3之间的2节点，可以有如下几种表示

node1->next = node3;
node1->next = node2->next;
node1->next = node1->next->next;

还可以这样表示：

ListNode **del = &node2;
*del = (*del)->next;

解决方法一

struct  ListNode
{
int val;
ListNode *next;
ListNode(int x):val(x),next(NULL){}
};

class Solution
{
public:
ListNode * removeNthFromEnd(ListNode* head, int n)
{
ListNode **del = &head, *iter = head;
for (int i = 0; i < n; ++i, iter = iter->next);
for (; iter != NULL; del = &((*del)->next), iter = iter->next);
*del = (*del)->next;
return head;
}
};

解决方法二：引入一个哑节点dummy，让快指针先走n+1步。

public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode dummy = new ListNode(0);
dummy.next = head;
ListNode first = dummy;
ListNode second = dummy;
for (int i = 1; i <= n + 1; i++) {
first = first.next;
}
while (first != null) {
first = first.next;
second = second.next;
}
second.next = second.next.next;
return dummy.next;
}