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.
Answer:
(1) head->1->2->3->4->5->6->7
p
q
(2) head->1->2->3->4->5->6->7 // move q, n elements next to p
p
q
(3) head->1->2->3->4->5->6->7 // q++ ,p++;
p
q
(4) head->1->2->3->4->5->6->7 // q++ ,p++;
p
q
(5) head->1->2->3->4->5->6->7 // q++ ,p++;
p
q
(6) head->1->2->3->4->5->6->7 // q++ ,p++;
p
q
(7) head->1->2->3->4----->6->7 // delete p->next
p
qreturn head->next;
/** * 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) { // Start typing your C/C++ solution below // DO NOT write int main() function ListNode *p= new ListNode(0); p->next=head; head = p; ListNode *q=p; int i=0; while (i<n){ q=q->next; i++; } while (q->next){ q=q->next; p=p->next; } p->next=p->next->next; return head->next; } };
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