【LeetCode100】--- 101.重排链表【思维导图+复习回顾】

题目传送门

方法一：（原地修改）

题目要求原地修改（不创建新链表节点，直接调整指针）进行合并链表

java 复制代码

class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return;
        }
        ListNode slow = head;
        ListNode fast = head;
//寻找链表中心节点
        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode head2 = reverseList(slow.next);
        slow.next = null;
        
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        boolean flag = true;

        ListNode h1 = head;
        ListNode h2 = head2;
//合并链表
        while(h1 != null && h2 != null){
            ListNode next1 = h1.next; // 保存h1的下一个节点
            ListNode next2 = h2.next; // 保存h2的下一个节点

            h1.next = h2; // h1连接h2
            h2.next = next1; // h2连接h1的原下一个节点

            h1 = next1; // 移动h1指针
            h2 = next2; // 移动h2指针
        }
    }
//反转链表
    public ListNode reverseList(ListNode slow){
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        
        while(slow != null){
            ListNode next = slow.next;
            slow.next = prev.next;
            prev.next = slow;
            slow = next;
        }
        return prev.next;
    }
}

方法二：（新建链表合并链表）

java 复制代码

/**
 * 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 void reorderList(ListNode head) {
        if(head == null || head.next == null){
            return;
        }

        ListNode slow = head;
        ListNode fast = head;

        while(fast.next != null && fast.next.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        ListNode head2 = reverseList(slow.next);
        slow.next = null;
        
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;

        boolean flag = true;

        ListNode h1 = head;
        ListNode h2 = head2;
        while(h1 != null || h2 != null){
            if(flag == true && h1 != null){
                prev.next = new ListNode(h1.val);
                h1 = h1.next;
                prev = prev.next;
                flag = false;
            }else if(h2 != null){
                prev.next = new ListNode(h2.val);
                h2 = h2.next;
                prev = prev.next;
                flag = true;
            }
        }
        //head = dummy.next;这样不对是因为
        //两个引用彻底指向不同对象，修改head的指向和originalHead无关
        head.next = dummy.next.next;
    }

    public ListNode reverseList(ListNode slow){
        ListNode dummy = new ListNode(0);
        ListNode prev = dummy;
        
        while(slow != null){
            ListNode next = slow.next;
            slow.next = prev.next;
            prev.next = slow;
            slow = next;
        }
        return prev.next;
    }
}