力扣：19.删除链表的倒数第 N 个结点

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode *dummy = new ListNode( 0 );
        dummy->next = head;
        ListNode *fast = dummy;
        ListNode *slow = dummy;
        for ( int i = 0; i < n + 1; i++ ) {
            fast = fast->next;
        }
        while ( fast != nullptr ) {
            fast = fast->next;
            slow = slow->next;
        }
        ListNode *del = slow->next;
        slow->next = slow->next->next;
        delete del;
        return dummy->next;
    }
};

1. 删除倒数第 N 个结点；
2. 先让 fast 领先 N + 1 个单位，当 fast 为空，slow 和 fast 中间隔了 N 个结点，slow 是待删除结点的前驱；
3. slow 的后继指向后继的后继，然后删除后继，释放内存。