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题目:
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.
首先容易想到的是先遍历一次链表,计算链表长度L,然后再次遍历到L-n的位置删除结点。更好的办法是应用递归,从后往前n递减,到-1时就到了要被删除结点的前一个结点。
/** * 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* node = new ListNode(0); node->next = head; removeNode(node, n); return node->next; } void removeNode(ListNode* head, int& n){ if (!head) return; removeNode(head->next, n); n--; if (n == -1) head->next = head->next->next; return; } };
【LeetCode】 19. Remove Nth Node From End of List 解题小结
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原文地址:http://www.cnblogs.com/Doctengineer/p/5848938.html