题解 | #删除链表的倒数第n个节点# C语言快慢指针

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 * };
 */
/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param head ListNode类 
 * @param n int整型 
 * @return ListNode类
 */
struct ListNode* removeNthFromEnd(struct ListNode* head, int n ) 
{
    struct ListNode* fast=head;
    struct ListNode* slow=head;
    struct ListNode* cur=head;
    //寻找倒数第n个结点-->slow
    while(n--)
    {
        fast =fast ->next;
    }
    while(fast)
    {
        fast =fast->next;
        slow =slow->next;
    }
    //如果slow为head，直接删除
    if(slow==head)
    {
        cur=slow->next;
        free(slow);
        return cur;
    }
    //当solw不为head时，寻找slow前一个结点
    while(cur->next != slow)
    {
        cur=cur->next;
    }
    cur->next=slow->next;
    free(slow);
    return head;
}