LeetCode-143 - Rearrange linked list

Rearrange linked list

Topic description: Given a head node head of a singly linked list L, the singly linked list L is expressed as:

L0 → L1 → … → Ln-1 → Ln
Please rearrange it to become:

L0 → Ln → L1 → Ln-1 → L2 → Ln-2 → …

You can't just change the value inside the node, but actually need to exchange the node.

For example descriptions, please refer to the official website of LeetCode.

Source: LeetCode
Link: https://leetcode-cn.com/problems/reorder-list/
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Solution 1: Linked List Traversal

First, if the linked list is empty or the linked list has only one node, return it directly.

Otherwise, first use a stack nodes to record all nodes, and record the number of linked list nodes count;

Then, record the insertion order. When traversing to the odd bit, insert it into the linked list from the direction of the head node; when traversing to the even bit, remove the node from the stack (that is, from the direction of the tail node) and insert it into the linked list.

import com.kaesar.leetcode.ListNode;

import java.util.Stack;

public class LeetCode_143 {
    public static void reorderList(ListNode head) {
        if (head == null || head.next == null) {
            return;
        }
        // 所有节点
        Stack<ListNode> nodes = new Stack<>();
        // 链表节点的数量
        int count = 0;
        ListNode cur = head;
        while (cur != null) {
            count++;
            nodes.push(cur);
            cur = cur.next;
        }

        int front = 1, back = 0, i = 1;
        ListNode newCur = head;
        cur = head.next;
        // 分别从头结点和栈中遍历链表节点，然后按指定顺序插入新的头节点构成的链表中
        while (front + back < count) {
            i++;
            if (i % 2 == 1) {
                // 插入正向的节点
                newCur.next = cur;
                cur = cur.next;
                front++;
            } else {
                // 插入后面的节点
                newCur.next = nodes.pop();
                back++;
            }
            newCur = newCur.next;
        }
        // 最后，要将新的尾结点的next指向null
        newCur.next = null;
    }

    public static void main(String[] args) {
        ListNode head = new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        head.next.next.next = new ListNode(4);
        head.next.next.next.next = new ListNode(5);

        System.out.println("-----重排之前-----");
        ListNode cur = head;
        while (cur != null) {
            System.out.print(cur.val + " ");
            cur = cur.next;
        }
        System.out.println();

        reorderList(head);
        System.out.println("-----重排之后-----");
        cur = head;
        while (cur != null) {
            System.out.print(cur.val + " ");
            cur = cur.next;
        }

    }
}

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