首页 > 代码库 > [LeetCode] Remove Nth Node From End of List
[LeetCode] Remove Nth Node From End of List
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.
Note: Given n will always be valid. Try to do this in one pass.
增加一个头结点，解决有可能需要删除原head问题，就不需要单独处理了。
/** * 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 *h = new ListNode(0); h->next = head; ListNode *p1 = h,*p2 = h; int num = 0; while(num!=n){ p2 = p2->next; num++; } while(p2->next != NULL){ p1 = p1->next; p2 = p2->next; } ListNode *p = p1->next->next; p1->next = p; return h->next; }
[LeetCode] Remove Nth Node From End of List
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。