本题解法代码的思想及编写参考了网址https://github.com/soulmachine/leetcode#leetcode题解
题目：Given a linked list, remove the nth node from the end of list and return its head. For example, Given linked list: 1->2->3->4->5, and n = 2. After removing the second node from the end, the linked list becomes 1->2->3->5. Note: • Given n will always be valid. • Try to do this in one pass
解析：设两个指针 cur 和 prev, 让 cur 先走 n 步，然后 prev 和 cur 一起走，直到 cur 走到尾节点，删除 prev -> next 即可
解题代码如下：
// 时间复杂度O(n)，空间复杂度O(1) class Solution { public: ListNode* removeNthFromEnd(ListNode* head, int n) { ListNode dummy{-1}; dummy.next = head; ListNode *prev = &dummy, *cur = &dummy; for (int i = 0; i != n; ++i) // cur 先走 n 步 cur = cur -> next; while (cur -> next) { // prev 与 cur 一起走 prev = prev -> next; cur = cur -> next; } ListNode* tmp = prev -> next; prev -> next = prev -> next -> next; delete tmp; return dummy.next } };