143. 重排链表

 解题过程

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if(head == null){
            return;
        }
        ListNode mid = middleNode(head);
        ListNode l1 = head;
        ListNode l2 = mid.next;
        mid.next = null;
        l2 = reverse(l2);
        merge(l1,l2);
    }

    // 找中点
    public ListNode middleNode(ListNode head){
        ListNode slow = head;
        ListNode fast = head;
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        return slow;
    }

    // 反转链表
    public ListNode reverse(ListNode head){
        ListNode pre = null;
        ListNode cur = head;
        while(cur != null){
            ListNode temp = cur.next;
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }

    // 链表两端合并
    public void merge(ListNode l1, ListNode l2){
        ListNode temp1;
        ListNode temp2;
        while(l1 != null && l2 != null){
            temp1 = l1.next;
            temp2 = l2.next;

            l1.next = l2;
            l1 = temp1;

            l2.next = l1;
            l2 = temp2;
        }
    }
}