LeetCode -- Remove Nth Node From End of List C 语言 AC code

Given the head of a linked list, remove the n<sup>th</sup> node from the end of the list and return its head.

Follow up: Could you do this in one pass?

Example 1:

image

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

哈哈哈哈 自己写出来的，用的还是递归的方式，感觉自己很棒啊！
这道题只是删除某个指定的结点，那么就可以使用递归的方式来计算链表的长度，然后和当前长度进行比较，如果达到了指定的节点那么就能够进行删除动作。不过感觉耗时比较多。。。不知咋回事。、

常规解法感觉没啥亮点，这里就不讨论了。

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

typedef struct ListNode ListNode;

struct ListNode* RecurseDeepIn(struct ListNode* node, int count, int n, int* depth){
    
    if( node != NULL ) node->next = RecurseDeepIn(node->next, count, ++n, depth);
    if( node == NULL) *depth = n;
    if( *depth != 0 && (*depth - n) == count-1) return node == NULL ? node : node->next;
    
    return node;
}

struct ListNode* removeNthFromEnd(struct ListNode* head, int n){
    int *depth;
    int depthCount = 0;
    depth = &depthCount;
    head = RecurseDeepIn(head, n, 0, depth);
    return head;
}