[LeetCode] Reorder List

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} .

解题思路1

首先用一个vector存储所有的结点，然后将倒数第i个结点插入到第i个结点之后。

实现代码1

//Runtime:82 ms
#include <iostream>
#include <vector>
using namespace std;

struct ListNode
{
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(NULL){}
};

class Solution {
public:
    void reorderList(ListNode *head) {
        if (head == NULL)
        {
            return;
        }
        ListNode *p = head;
        vector<ListNode*> nodes;
        while (p)
        {
            nodes.push_back(p);
            p = p->next;
        }

        int left = 0;
        int right = nodes.size() - 1;
        while (left < right)
        {
            nodes[left]->next = nodes[right];
            nodes[right--]->next = nodes[++left];
        }
        nodes[left]->next = NULL;
    }
};
int main()
{
    ListNode *p1 = new ListNode(1);
    ListNode *p2 = new ListNode(2);
    ListNode *p3 = new ListNode(3);
    ListNode *p4 = new ListNode(4);
    p1->next = p2;
    p2->next = p3;
    p3->next = p4;
    Solution s;
    s.reorderList(p1);

    ListNode *p = p1;
    while (p)
    {
        cout<<p->val<<endl;
        p = p->next;
    }
}

解题思路2

首先把链表从中间分成两部分，然后将后一部分链表反转，最后将两个链表合并。

实现代码2

//Runtime:98 ms
class Solution {
public:
    void reorderList(ListNode *head) {
        if (head == NULL)
        {
            return;
        }
        ListNode* ps = head;
        ListNode* pf = head;
        while (pf->next && pf->next->next)
        {
            ps = ps->next;
            pf = pf->next->next;
        }

        ListNode *p2 = ps->next;
        ps->next = NULL;
        p2 = reverseList(p2);
        head = mergeList(head, p2);
    }

    ListNode* mergeList(ListNode *p1, ListNode *p2)
    {
        ListNode *result = p1;
        ListNode *temp;
        while (p2)
        {
            temp = p2;
            p2 = p2->next;

            temp->next = p1->next;
            p1->next = temp;
            p1 = p1->next->next;
        }

        return result;
    }

    ListNode* reverseList(ListNode *head){
        if (head == NULL)
        {
            return NULL;
        }
        ListNode *p = head->next;
        head->next = NULL;
        while (p)
        {
            ListNode *temp = p;
            p = p->next;

            temp->next = head;
            head = temp;
        }

        return head;
    }
};