标签：init lse code public rem == example lis and
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?
/** * 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* p=head; int len=0; while(p!=NULL) { len++; p=p->next; } p=head; for(int i=1;i<len-n;i++) p=p->next; if(len-n==0) { head=head->next; return head; } ListNode* q=p->next; if(p->next!=NULL) p->next=p->next->next; else p->next=NULL; delete q; return head; } };
19. Remove Nth Node From End of List
标签：init lse code public rem == example lis and
原文地址：https://www.cnblogs.com/guoshuai1995/p/10408601.html
