[LeetCode] Reorder List

1.Description

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}.

2.Analysis

To solve this problem, we have two attention Note.

A. The pointer exchange order.

             Assume List {1 2 3 4  5}

            pointer    p1 L0  1

                            p2  Ln  5

                            p3  Ln-1 4

             the reorder should be 1 5 2 3 4

             the opertation should   p3 = NULL;

                                                       p2.next = p1.next;

                                                        p1.next = p2;

             draw a picture about the order can help greatly.

B. Time

        We draw the conclusion we need to search the List n/2 times for the reorder the computation complexity should be O(n^2)

        This indeed waste much time.

        For the reorder List, L0→Ln→L1→Ln-1→L2→Ln-2→…

        We need the location of Ln, Ln-1, Ln-2.....This means if we know the their location  prior to reorder, we don't need to search the list again and again.

        Solution:

        Search the List and record their Location.

NOTE: This can be treated as the reducing the operation time at the expense of using more space.

3, CODE:

     Time Exceed Code:

      

if(head == NULL || head->next == NULL)        return;    ListNode *p1, *p2, *p3;    std::vector<ListNode*> location;    std::vector<ListNode*>::iterator it;    p1 = head;    while(p1 != NULL)    {        location.insert(location.begin(), p1);        p1 = p1->next;    }    p1 = head;    it = location.begin();    p2 = *(it++);    p3 = *(it);    while(it!= location.end() && p2 != p1 && p2 != p1->next)    {        p3->next = NULL;        p2->next = p1->next;        p1->next = p2;        p1 = p2->next;        p2 = *(it++);        p3 = *(it);    }

AC code

if(head == NULL || head->next == NULL)            return;    ListNode *p1, *p2, *p3;    std::vector<ListNode*> location;    std::vector<ListNode*>::iterator it;    p1 = head;    while(p1 != NULL)    {        location.insert(location.begin(), p1);        p1 = p1->next;    }    p1 = head;    it = location.begin();    p2 = *(it++);    p3 = *(it);    while(it!= location.end() && p2 != p1 && p2 != p1->next)    {        p3->next = NULL;        p2->next = p1->next;        p1->next = p2;        p1 = p2->next;        p2 = *(it++);        p3 = *(it);    }