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

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