Given a linked list, remove the $n_{th}$ node from the end of list and return its head.

题目链接

19. Remove Nth Node From End of List

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.

代码

class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        //三指针
        ListNode *pointa = head;//b前边的那个
        ListNode *pointb = head;
        ListNode *pointc = head; 
        if (head == NULL || n < 1) return head;
        //前边指针先走n步
        for (int i = 0; i < n; i++) {
            pointc = pointc->next;
            if (pointc == NULL&&i == n - 1) return head->next;
            else if (pointc == NULL) return head;
        }

        while (pointc != NULL) {
            pointc = pointc->next;
            pointa = pointb;
            pointb = pointb->next;
        }

        pointa->next = pointb->next;

        return head;
    }
};

一次就ac了，还有点惊讶。。。

本文由 mmmwhy 创作，最后编辑时间为: Jun 6, 2018 at 02:55 pm