刷题——ReorderList

/* * 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}. */public class ReorderList {//快慢指针找中间节点public ListNode findCenter(ListNode head){if (head==null || head.next==null) {return head;}ListNode slowListNode = head;ListNode fastListNode = head;while (fastListNode!=null && fastListNode.next!=null) {slowListNode = slowListNode.next;fastListNode = fastListNode.next.next;}return slowListNode;}//链表反转public ListNode reverseList(ListNode head){if (head==null || head.next==null) {return head;}//利用首插法进行反转ListNode dummy = new ListNode(-1);dummy.next = head;while (head.next!=null) {ListNode nextNode = head.next;head.next = nextNode.next;nextNode.next = dummy.next;dummy.next = nextNode;}return dummy.next;}//两个链表进行合并public void mergeList(ListNode head,ListNode head2){if (head==null) return;ListNode tailListNode = head;while (head2!=null) {ListNode nextListNode = head2.next;//短链中头的下一个节点head2.next = tailListNode.next;tailListNode.next = head2;head2 = nextListNode;tailListNode = tailListNode.next.next;}}public void reorderList(ListNode head) { if (head==null || head.next==null) {  return; }     ListNode centerListNode = findCenter(head);//找中间节点     ListNode nextListNode = centerListNode.next;     centerListNode.next = null;//断开链     ListNode head2 = reverseList(nextListNode);//反转后面链表     mergeList(head,head2);//合并两个链表     }public static void main(String[] args) {// TODO Auto-generated method stubListNode listNode0 = new ListNode(1);ListNode listNode1 = new ListNode(2);ListNode listNode2 = new ListNode(3);ListNode listNode3 = new ListNode(4);ListNode listNode4 = new ListNode(5);listNode0.next = listNode1;listNode1.next = listNode2;listNode2.next = listNode3;listNode3.next = listNode4;new ReorderList().reorderList(listNode0);while (listNode0!=null) {System.out.print(listNode0.val+" ");listNode0 = listNode0.next;}System.out.println();}}