标签：for bsp ring linked val out with this public
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}.
class Solution {
public void reorderList(ListNode head) {
if (head == null || head.next == null)
return;
//找到链表的中点，如1 2 3 4 5 6 7 8，则p1是4
ListNode p1 = head;
ListNode p2 = head;
while (p2.next != null && p2.next.next != null) {
p1 = p1.next;
p2 = p2.next.next;
}
//链表后半边逆序，5 6 7 8 变成 8 7 6 5
ListNode preMiddle = p1; //preMiddle 是4
ListNode preCur = p1.next; //preCur 是5
while (preCur.next != null) {
ListNode cur = preCur.next; //cur是6
preCur.next = cur.next; //5->7
cur.next = preMiddle.next; //6->5
preMiddle.next = cur; //4->6
}
//达到最后排序目的
p1 = head;
p2 = preMiddle.next;
while (p1 != preMiddle) {
preMiddle.next = p2.next; //另4->7
p2.next = p1.next; //8->2
p1.next = p2;//1->8
p1 = p2.next;//让p1等于2
p2 = preMiddle.next; //p2等于7，现在链表1 8 2 3 4 7 6 5
}
}
}
143. Reorder List
标签：for bsp ring linked val out with this public
原文地址：https://www.cnblogs.com/MarkLeeBYR/p/10677654.html
