Given a linked list, remove the nth node from the end of list and return its head.
For 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
题目要求在一个链表里删除倒数第n个结点。使用两个指针，第一个指向第n个节点，第二个指向头节点。再往后遍历直到第一个指针指向尾节点，则此时第二个指针指向倒数第n个结点的前一个节点。删除第二个指针节点。 /** * 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) { struct ListNode *p=head,*q=head; for(int i = 0;i<n;i++) { p = p->next; if(!p) return head->next; } while(p->next) { q = q->next; p = p->next; } q->next = q->next->next; return head; } };