https://leetcode.com/problems/remove-nth-node-from-end-of-list/
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) { } };
/** * 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* head1 = head; ListNode* head2 = head; for (int i = 0; i < n; i ++){ head2 = head2->next; } if (head2 == NULL) return head->next; // remove the first node while (head2->next){ head2 = head2->next; head1 = head1->next; } head1->next =head1->next->next; return head; } };