[Leetcode] 019. Remove Nth Node from End of List (Easy) (C++/python)

Index: [Leetcode] leetcode key index (C++/JAVA/PYTHON/SQL)
Github:https://github.com/illuz/leetcode

019.remove_nth_node_from_end_of_list (Easy) links：

Title: https://oj.leetcode.com/problems/remove-nth-node-from-end-of-list/
Code (GitHub): Https://github.com/illuz/leetcode

Test Instructions：

Deletes the reciprocal nth node of a one-way linked list.

Analysis：

1. Direct simulation, first calculate the number of nodes, and then find the node delete
2. With two pointers, one goes first N steps, and then goes together.

Here in C + + implementation of the first, with Python implementation of the second.
Java words and C++/python almost, do not write out.

Code：

C++:

Class Solution {public:    listnode *removenthfromend (listnode *head, int n) {if (n = = 0) return head;//count the node Nu Mberint num = 0; ListNode *cur = head;while (cur! = NULL) {cur = cur->next;num++;} if (num = = N) {//Remove first Nodelistnode *ret = Head->next;delete head;return ret;} else {//remove (cnt-n) th Nodein T m = num-n-1;cur = Head;while (m--) cur = cur->next; ListNode *rem = Cur->next;cur->next = Cur->next->next;delete rem;return head;}}    ;

Python:

Class solution:    # @return a listnode    def removenthfromend (self, head, N):        dummy = listnode (0)        Dummy.next = head        p, q = dummy, dummy                # first ' Q ' Go n step for        I in range (n):            q = q.next        # Q & p< C9/>while q.next:            p = p.next            q = q.next        rec = p.next        p.next = rec.next        del rec        return dummy . Next

[Leetcode] 019. Remove Nth Node from End of List (Easy) (C++/python)