19. Remove Nth Node From End of ListV-优快云博客

本文探讨了一次性遍历链表并删除倒数第N个节点的算法，通过使用快慢指针技巧，确保慢指针恰好位于目标节点前一个位置，从而实现高效操作。

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

这个题目考察的是一次性遍历,如果要一次性就必须要用指针,快慢指针是最常见的方法,这个题目就是要使得慢指针刚好指向要删除目标节点的前一个,让快指针指向最后,与慢指针相差n步

/**
 * 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) {
        vector<ListNode*> res;
        ListNode* resNode = new ListNode(0);
        ListNode* slow = resNode;
        ListNode* fast = resNode;
        resNode->next = head;
        for(int i=0;i<=n;i++){
            fast = fast->next;
        }
        while(fast!=NULL){
            slow = slow->next;
            fast = fast->next;
        }
        slow->next = slow->next->next;
        return resNode->next;
    }
};```