题解 | #重排链表#

/**
 * 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) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode slow = dummy;
        ListNode fast = dummy;
        while(fast!=null && fast.next!=null){//此时还没有到达中点位置
            slow = slow.next;
            fast = fast.next;
            if(fast.next!=null){
                fast = fast.next;
            }
        }
        ListNode tmp = slow.next;//中点后的链表
        slow.next = null;
        link(head,reverseList(tmp),dummy);

    }
    //将两个链表连接起来
    public void link(ListNode node1,ListNode node2,ListNode head){
        ListNode cur = head;
        while(node2!=null){ //构建链表
            ListNode tmp = node1.next;
            cur.next = node1;
            node1.next = node2;
            cur = node2;
            node1 = tmp;
            node2 = node2.next;
        }
        if(node1!=null){
            cur.next = node1;
        }
    }
    //反转链表
    public ListNode reverseList(ListNode node){
        ListNode pre = null;
        ListNode cur = node;
        while(cur!=null){
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;

    }
}