33：Remove Nth Node From End of List

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
        }
};