19. Remove Nth Node From End of List

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?

2级指针删除, 避免了prev变量, 也不需要判断是否为头指针的问题,也不用增加一个额外的head指针

先用l1 移动n步, 接着l1 和 l2 一起移动直到l1 到末尾

/**
 * 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) {
        ListNode *l1=head, **l2=&head;
        for(int i=1;i<n;++i) //这里循环n-1次即可
            l1=l1->next;
        while(l1->next) //注意l2的初始值是&head, 所以第一次循环后 l2的值为  &head->next, 接下来为 & head->next->next
        {
            l1=l1->next;
            l2=&(*l2)->next;
        }
        *l2=(*l2)->next; //此时l2指向需要删除的前一个节点的next, 所以直接赋值为下一个next即可
        return head;
    }
};