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.
Solution in C++
/** * 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) { vector<ListNode *> e; ListNode *p=head; while(p) { e.push_back(p); p=p->next; } if (n==e.size()) return head->next; else if (n==1) e[e.size()-2] -> next = NULL; else e[e.size()-n-1]->next = e[e.size()-n]->next; return head; } };
Solution in Java - Time Complexity: O(L), Space Complexity: O(1)
public ListNode removeNthFromEnd(ListNode head, int n) { ListNode dummy = new ListNode(0); dummy.next = head; int length = 0; ListNode first = head; while (first != null) { length++; first = first.next; } length -= n; first = dummy; while (length > 0) { length--; first = first.next; } first.next = first.next.next; return dummy.next; }