LeetCode-19-Remove Nth Node From End of List

算法描述：

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?

 

解题思路：

这个题考查数据结构。设置两个指针，快指针先走n步，然后快指针和慢指针一起走，快指针停的时候，慢指针的下一个元素为待删除元素。

    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* dup = new ListNode(-1);
        dup->next = head;
        ListNode* fast = dup;
        while(n>0){
            fast=fast->next;
            n--;
        }
        ListNode* slow = dup;
        while(fast!=nullptr && fast->next!=nullptr){
            fast=fast->next;
            slow=slow->next;
        }
        slow->next = slow->next->next;
        return dup->next;
    }