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LeetCode[Linked List]: Remove Nth Node From End of List

时间:2014-11-03 21:02:22      阅读:182      评论:0      收藏:0      [点我收藏+]

标签:leetcode   linked list   nth node   空间复杂度   算法   

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.
Note:
Given n will always be valid.
Try to do this in one pass.

这个题目我想到两种解法。时间复杂度均为O(N),空间复杂度有所区别,一种是O(N),一种是O(1)

另外,这个题目我觉得需要注意的地方是应该delete掉删除的节点,防止内存泄露。

空间复杂度为 O(N) 的解法

需要用到一个vector保存所有节点的地址,具体代码如下:

C++ code
ListNode *removeNthFromEnd(ListNode *head, int n) { vector<ListNode *> list; for (ListNode *iter = head; iter != NULL; iter = iter->next) list.push_back (iter); if (n == list.size()) { head = list[0]->next; delete list[0]; return head; } list[list.size() - n - 1]->next = list[list.size() - n]->next; delete list[list.size() - n]; return list[0]; }

空间复杂度为 O(1) 的解法

设置两个指针,fast较low快n个节点。具体代码如下:

C++ code
ListNode *removeNthFromEnd(ListNode *head, int n) { ListNode *fast = head, *low = head, *delNode; for (int i = 0; i < n; ++i) fast = fast->next; if (!fast) { delNode = head; head = head->next; delete delNode; return head; } while (fast->next) { fast = fast->next; low = low ->next; } delNode = low->next; low->next = low->next->next; delete delNode; return head; }

LeetCode[Linked List]: Remove Nth Node From End of List

标签:leetcode   linked list   nth node   空间复杂度   算法   

原文地址:http://blog.csdn.net/chfe007/article/details/40745917

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