309. 19. Remove Nth Node From End of List[Medium]

Description

Given a linked list, remove the nth 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.
Try to do this in one pass.

Solution

挺简单的一道题。删除链表哦中倒数第n个数，要求只能遍历一次，所以不能先遍历一次看共有多少个元素再重新遍历到size-n。
用两个指针，一个指针先走n步，此时如果把尾看成头，则相当于走到了倒数第n+1个数的位置。另一个指针从这时候开始，两个指针同时移动，当第一个指针移动到尾节点的时候，第二个指针的next就是要被删除的节点。
需要注意的是，如果第一个指针走完n步已经到尾节点，说明删除的就是head，返回head->next即可。

Complexity analysis

O(n)

Code

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head -> next == NULL || head == NULL)
            return NULL;
        ListNode* p = head;
        for (int i = 0; i < n; i++) {
            p = p -> next;
        }
        if (p == NULL)
            return head -> next;
        else {
            ListNode* q = head;
            while(p->next != NULL) {
                q = q -> next;
                p = p -> next;
            }
            ListNode* temp = q -> next;
            q -> next = temp -> next;
            delete temp;
        }
        return head;
    }
};

Result