Leetcode--Reorder List

Problem 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}.

分析：按照要求交换链表的元素，第一种方法，直接遍历一遍链表将所有节点指针保存在vector中，然后一头一尾不断重建链表即可。具体代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void reorderList(ListNode *head) {        ListNode *p=head;        int length,i=1,tag=0;        vector<ListNode* > array;        while(p)        {            array.push_back(p);            p=p->next;        }        p=head;        if(array.size()<3) return;        length=array.size()-1;        while(tag<array.size()/2)        {            p->next=array[length--];            p=p->next;            p->next=array[i++];            p=p->next;            tag++;        }        p->next=NULL;    }};

第二种方法就是找到链表的中间节点，将链表分为两半，然后将后半部分转置，最后将前后两部分进行合并即可，具体代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    void reorderList(ListNode *head) {        if (!head){return;}        if (head->next==NULL){return;}        ListNode *p=head;         ListNode *q=head;                  //find the midddle pointer        while (q->next && q->next->next){            p=p->next;            q=q->next->next;        }                 //now p is middle pointer        //reverse p->next to end        q = p->next;        while (q->next){            ListNode* tmp = p->next;            p->next = q->next;            q->next = q->next->next;            p->next->next = tmp;        }                 //reorder        q = head;        while (p!=q && p->next){            ListNode* tmp = q->next;            q->next = p->next;            p->next = p->next->next;            q->next->next = tmp;            q=q->next->next;        }        return;    }};