文章目录

思路

双指针,首先让快指针先走n个点.这时候和slow的距离便是从最后一个点到要删的点的距离

解题方法

定义一个fast指针,和一个slow指针,指向dummyhead(虚头结点,便于统一操作), 首先让快指针先走n个点.这时候和slow的距离便是从最后一个点到要删的点的距离

复杂度

• 时间复杂度:
  • O ( n )
• 空间复杂度:
  • O ( 1 )

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummyHead = new ListNode(0);
        dummyHead.next = head;
        ListNode fastIndex = dummyHead;
        ListNode slowIndex = dummyHead;  
        n++; //让fast指针多走一步,从而让slow->next指向被删的点
        while(n-- != 0&& fastIndex!=null){
            fastIndex = fastIndex.next;
        }
        while(fastIndex != null){
            slowIndex = slowIndex.next;
            fastIndex = fastIndex.next;
        }
        slowIndex.next = slowIndex.next.next;
        return dummyHead.next;  
    }
}

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode* fast = dummyHead; // 快指针
        ListNode* slow = dummyHead;
        n++; //因为要操作的是slow->next,同时也为了避免fast为空指针的情况
        while(n-- && fast != NULL){
            fast = fast->next;
        }

        while(fast != NULL){
            fast = fast->next;
            slow = slow->next;
        }
        slow->next = slow->next->next;
        return dummyHead->next;
    }
};