ama-non

Given a linked list, remove the n-th node from the end of list and return its head.

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.

Follow up:

Could you do this in one pass?

分析：

设两个指针p,q，让q先走n步，然后p和q一起走，直到q走到尾节点，删除p->next即可。

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
        // ListNode dummy{-1, head}; (此处为C11标准列表初始化)
        ListNode dummy = ListNode(0);
        dummy.next = head;
        ListNode *p = &dummy, *q = &dummy;

        for (int i = 0; i < n; i++)  // q先走n步
            q = q->next;

        while(q->next != nullptr) { // 一起走
            p = p->next;
            q = q->next;
        }
        ListNode *tmp = p->next;
        p->next = p->next->next;
        delete tmp;
        return dummy.next;
    }
};

时间复杂度o(n)，空间复杂度o(1)

参考来源：

• https://soulmachine.gitbooks.io/algorithm-essentials/content/cpp/linear-list/linked-list/remove-nth-node-from-end-of-list.html