LeetCode (力扣) 19. Remove Nth Node From End of List (C) - Medium

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题目为给定一个链表 ( Linked List ) 与一个数字 n ，n 代表要移除从后面数来第几个节点。

 

题目与范例如下

 

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

Example 1:

Input: head = [1,2,3,4,5], n = 2
Output: [1,2,3,5]
Example 2:

Input: head = [1], n = 1
Output: []
Example 3:

Input: head = [1,2], n = 1
Output: [1]

Constraints:

The number of nodes in the list is sz.
1 <= sz <= 30
0 <= Node.val <= 100
1 <= n <= sz

Follow up: Could you do this in one pass?

---

 

解题策略为透过三个指针，第一个为遍历整个链表( Link List ) 用的，第二个指针为与第一个指针差n ，用以指出倒数第n 个节点，第三个指针为第二个指针后面一个，用以移除第n 个节点用。

 

下方为我的代码

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
    struct ListNode* first = head;
    struct ListNode* second = head;
    struct ListNode* secondback = head;
    for(int i = 0;i<n;i++){
        first = first->next;
    }
    while(first!=NULL){
        if(second == secondback){
        }
        else{
            secondback = secondback->next;
        }
        first = first->next;
        second = second->next;
    }
    if(second == head)
        head = second->next;
    else
        secondback->next = second->next;
    
    return head;
}

 

下方为时间与空间之消耗

Runtime: 0 ms, faster than 100% of C online submissions for Remove Nth Node From End of List.

Memory Usage: 5.8 MB, less than 81.55% of C online submissions for Remove Nth Node From End of List.