[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 msclass 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;    }};