leetcode143. Reorder List

medium程度题

题目:

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You must do this in-place without altering the nodes' values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.

我想的是把n个节点存入容器，然后遍历容器改造节点，时间复杂度O(N)，空间复杂度O（N）

AC解:

class Solution {public:    void reorderList(ListNode* head)    {        if (head == nullptr || head->next == nullptr)            return ;        vector<ListNode*> vec;        //遍历一遍链表,所有节点存入vec        ListNode *cur = head;        while (cur)        {            vec.push_back(cur);            cur = cur->next;        }        //改造节点        auto first = vec.begin();        auto second = vec.end() - 1;        while (first < second)        {            (*first)->next = *second;            first++;            (*second)->next = *first;            second--;        }                (*first)->next = nullptr;    }};

另有一种空间复杂度O(1)的解法，把链表拆分成两个，将后面的链表翻转，然后合并两个链表

AC解:

class Solution {public:    void reorderList(ListNode* head)    {        if (head == nullptr || head->next == nullptr)            return ;        ListNode *fast = head,*slow = head,*prev = nullptr;        //拆分链表        while (fast && fast->next)        {            prev = slow;            slow = slow->next;            fast = fast->next->next;        }                prev->next = nullptr;        slow = reverse(slow);        //合并两个链表        ListNode *cur = head;        while (cur->next)        {            ListNode *temp = cur->next;            cur->next = slow;            slow = slow->next;            cur->next->next = temp;            cur = temp;        }                cur->next = slow;    }        ListNode* reverse(ListNode *head)    {        if (head == nullptr || head->next == nullptr)            return head;        ListNode *prev = head;        for (ListNode *cur = head->next,*next = cur->next; cur;)        {            head->next = next;            cur->next = prev;            prev = cur;            cur = next;                        next = next ? next->next : nullptr;        }                return prev;    }};

两种解法时间性能上没有差别