Reorder List (Java)

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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}.

把找到的中位数的下一个结点及之后的链表逆序，然后插入到前面的链表中。

Source

/** * Definition for singly-linked list. * class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { *         val = x; *         next = null; *     } * } */public class Solution {    public void reorderList(ListNode head) {    if(head == null || head.next == null) return ;   //***        ListNode p = head, q = head;        while(q.next != null && q.next.next != null){//遍历一遍找到中间结点的方法就是一个指针走 一步另一个走两步        p = p.next;        q = q.next.next;        }        q = p.next;        p.next = null; //***                ListNode ppre = null;        ListNode ppost;    //ppre指向p的前一个结点 ppost指向p的下一个结点        while(q != null){//链表逆置就是把链表指针顺序换个方向即可         ppost = q.next;        q.next = ppre;        ppre = q;        q = ppost;                }        q = ppre;   //此时是最后一个结点是头结点        p = head;        ListNode temp1, temp2;        while(p != null && q != null){        temp1 = p.next;        temp2 = q.next;        q.next = temp1;        p.next = q;        p = temp1;        q = temp2;        }    }}

Test

    public static void main(String[] args){    ListNode a = new ListNode(1);    a.next = new ListNode(2);    a.next.next = new ListNode(3);    a.next.next.next = new ListNode(4);     a.next.next.next.next = new ListNode(5);        new Solution().reorderList(a);    while(a != null){    System.out.println(a.val);    a = a.next;    }    }

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