Leetcode19. 删除链表的倒数第 N 个结点

 

 

19. 删除链表的倒数第 N 个结点

自己纯手写第一题，递归有点冗杂，开辟了虚拟头节点，而且要特别注意边界条件（当倒数第n个正好是头节点时）。

**
 * 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:
    int recursion(ListNode* node,ListNode* &head, int n){
        if(node->next == nullptr){
            int cnt = 0;
            return cnt;
        }
        int cnt = recursion(node->next, head, n);
        cnt ++;
        if(cnt == n){
            if(node->next == head){
                head = head->next;
            }
            else{
                node->next = node->next->next;
            }
        }  
        return cnt;
    }

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *node = new ListNode(-1);
        node->next = head;
        recursion(node, head, n);
        return head;
    }
};

大佬的代码永远是那么简洁利落

/**
 * 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:
    int cur = 0;
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if(head == nullptr) return nullptr;
        head->next = removeNthFromEnd(head->next, n);
        cur ++;
        if(cur == n) return head->next;
        return head;
    }
};

大佬的时间0ms，我的递归4ms，但是我没看出来是由于哪个原因导致时间花费是数倍，感谢如果有好心大佬在评论区能指导一下。

最经典的解法莫过于双指针了。

/**
 * 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* first = head;
        ListNode* second = head;

        if(head == nullptr) return nullptr;
        int i = 0;
        while(i ++ < n){
            first = first -> next;//快指针先走n步
        }
        if(first == nullptr) return head = head->next;//如果快指针走到头了，说明倒数第n个节点是头节点

        while(first->next != nullptr){
            first = first -> next;
            second = second -> next;
        }
        second ->next = second->next->next;
        return head;
    }
};