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.
/** * 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||n==0)return head; ListNode *p, *np; p = head; np = head; while(np) { np = np->next; n--; if(n==0) break; } if(!np&&n==0) return p->next; if(!np&&n!=0) return NULL; np = np->next; while(np) { p = p->next; np = np->next; } p->next = p->next->next; return head; } };代码略长, 边界状态慎重
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