【LeetCode】19. Remove Nth Node From End of List-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_27991659/article/details/50987601

本文介绍了一种高效算法，用于从链表中删除倒数第N个节点。通过使用快慢两个指针，该算法仅需遍历一次链表即可完成操作。文章提供了详细的实现代码，并考虑了特殊情况，如删除头结点的情况。

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.

两个指针相隔n-1，当一个指针走到末尾了，后一个指针就是要删除的节点。

/**
 * 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) {
      if (head == NULL ||n==0) return head;
       ListNode *tmp = NULL;
         ListNode *p = head;
         ListNode *q = head;
         for(int i = 0; i < n - 1; i++)
             q = q->next;
         while(q->next)
         {
             tmp = p;
             p = p->next;
             q = q->next;
         }
         if (tmp == NULL)
         {
             head = p->next;
             delete p;
         }
         else
         {
             tmp->next = p->next;
             delete p;
         }
         return head;
    }
};

二刷：同样是快慢指针。快指针先走N步，注意当删除的是头结点的情况，此时快指针已经到达链表末尾

/**
 * 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) {
        if(head==NULL || head->next==NULL || n <= 0) return NULL;
        ListNode *fast = head;
        ListNode *slow = head;
        ListNode *dummy = new ListNode(0);
        dummy->next = slow;
        for(int i = 0; i < n; i++)
        {
            fast  = fast->next;
        }
        if(fast==NULL)
        {
            dummy->next = slow->next;
        }
        while(fast && fast->next)
        {
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return dummy->next;
    }
};