剑指 Offer II 026. 重排链表

[题目链接](剑指 Offer II 026. 重排链表 - 力扣（LeetCode） (leetcode-cn.com))

思路

• 可以分成三个部分，1.将整个链表从中间进行分割成两个链表 2.将后半段链表进行翻转 3.将两段链表进行拼接
• 当链表节点数为奇数时，前半段链表会多出一个节点

代码

/**
 * 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 ListNode reverseNode(ListNode head) {//返回反转链表 //也会将原来的链表进行翻转
        ListNode pre = null;
        ListNode node = head;
        while(node != null) {
            ListNode next = node.next;
            node.next = pre;
            pre = node;
            node = next;
        }
        return pre;
    }
    private void link(ListNode node1, ListNode node2, ListNode head){//将分割的两段链表进行拼接
        ListNode pre = head;

        while(node1 != null && node2 != null) {
            ListNode next = node1.next;
            pre.next = node1;
            node1.next = node2;
            pre = node2;

            node2 = node2.next;
            node1 = next;
        }
        if(node1 != null) pre.next = node1; //如果链表的节点数是奇数个，那么第一段链表会多一个节点
    }
    public void reorderList(ListNode head) { //使用快慢双指针将链表从中间分割开
       ListNode dummy = new ListNode(0);
       dummy.next = head; //哨兵节点

       ListNode fast = head;
       ListNode slow = head;
       while(fast != null && fast.next != null){
           slow = slow.next;
           fast = fast.next;

           if(fast.next != null) fast = fast.next;
       }

       ListNode second = slow.next;
       slow.next = null;
       link(head,reverseNode(second),dummy);
    }
}