letcode改变链表的位置

最新推荐文章于 2025-01-18 19:25:59 发布

原创最新推荐文章于 2025-01-18 19:25:59 发布 · 155 阅读

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博客围绕LeetCode刷题中改变链表位置的问题展开。题目要求将单链表重新排序，且需原地操作不改变节点值。作者分享最初思路不够简洁，看到他人代码后反思，提出若有倒数第二个结点的操作方法，强调要发现问题最本质特征。

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letcode刷题

改变链表的位置

Given a singly linked list L: L 0→L 1→…→L n-1→L n,
reorder it to: L 0→L n →L 1→L n-1→L 2→L n-2→…

You must do this in-place without altering the nodes’ values.

For example,
Given{1,2,3,4}, reorder it to{1,4,2,3}.

这是自己最开始的想法，写的不够简洁。

class ListNode {
    int val;
    ListNode next;
    ListNode(int x) {
        val = x;
        next = null;
    }
}
public class Solution {
    public void reorderList(ListNode head) {
        if(head == null || head.next == null) {
            return;
        }
        ListNode nHead = new ListNode(0);
        ListNode cur = head;
        int i = 0;
        ListNode dump = nHead;
        while (cur != null) {
            if (i == 0) {
                ListNode iNode = cur;
                cur = cur.next;
                dump.next = iNode;
                iNode.next = null;
                dump = dump.next;
                i = 1;
            } else {
                ListNode lastNode = lastNode(cur);
                if (lastNode == cur) {
                    dump.next = lastNode;
                    break;
                } else {
                    ListNode preLastNode = nodeBeforeLast(cur, lastNode);
                    preLastNode.next = null;
                    dump.next = lastNode;
                    dump = dump.next;
                    i = 0;
                }
            }
        }
        head = nHead.next;
    }

    public ListNode lastNode(ListNode head) {
        if (head == null) {
            return null;
        }
        while (head.next != null) {
            head = head.next;
        }
        return head;
    }

    public ListNode nodeBeforeLast(ListNode head, ListNode last) {
        while (head.next != last) {
            head = head.next;
        }
        return head;
    }

    public static void main(String[] args) {
        ListNode head =  new ListNode(1);
        head.next = new ListNode(2);
        head.next.next = new ListNode(3);
        Solution solution = new Solution();
        solution.reorderList(head);
    }
}

再看了别人写的，感觉真的是几行代码的问题，然后自己就写了那么多行.如果有倒数第二个结点，就先找到倒数第二结点，倒数第二结点指向空，倒数第一个结点指向此时头结点的下一个结点，头节的下一个结点变成倒数第一个结点。然后头结点指向下下一个结点。

public class Solution {
    public void reorderList(ListNode head) {
        ListNode rear=head;
        ListNode tmp;
        while(head!=null&&head.next!=null){
            while(rear.next.next!=null){
                rear=rear.next;
            }
            tmp=rear.next;
            rear.next=null;
            tmp.next=head.next;
            head.next=tmp;
            head=rear=tmp.next;
        }
    }
   }