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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.
题目很简单,设置两个指针相隔n-1,一起走,走后一个指针走到头时,前一个指针正好在要删除的节点位置。要注意链表长度小于n的情况。
我的代码:
/** * 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 == NULL || n == 0) return head; ListNode *pf, *pl,*temp=NULL; pf = pl = head; int i = 0; for (; i < n-1; ++i){ if (pl->next) pl = pl->next; else return head; } while (pl->next){ temp = pf; pl = pl->next; pf = pf->next; } if (temp == NULL){ head = pf->next; delete pf; } else{ temp->next = pf->next; delete pf; } return head; } };
#19 Remove Nth Node From End of List
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原文地址:http://www.cnblogs.com/Scorpio989/p/4450913.html