Reorder List

最新推荐文章于 2022-10-08 18:49:12 发布

原创最新推荐文章于 2022-10-08 18:49:12 发布 · 372 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

66 篇文章

订阅专栏

本文详细阐述了如何通过快慢指针找到链表中点、反转后半部分链表，并将其与前半部分交替链接，实现链表的重组。通过实例演示了从原始链表到重组链表的完整过程。

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-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}.

先用快慢指针找到能把原list切分两半的中点，然后把后半段list 反转，最后把前半段list和反转后的后半段list交替链接起来就行了。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    void reorderList(ListNode *head) {
        if (head == NULL || head->next == NULL) return;
        ListNode *mo = head, *hi = head->next;
        while (hi != NULL && hi->next != NULL) {
            hi = hi->next->next;
            mo = mo->next;
        }
        
        ListNode *m1 = mo->next;
        mo->next = NULL;
        hi = reverseList(m1);
        mo = head;
        while (mo != NULL && hi != NULL) {
            m1 = mo->next;
            mo->next = hi;
            ListNode *h1 = hi->next;
            hi->next = m1;
            mo = m1;
            hi = h1;
        }
    }
private:
    ListNode *reverseList(ListNode *head) {
        if (head == NULL || head->next == NULL) return head;
        ListNode *pre = head;
        ListNode *cur = head->next;
        head->next = NULL;
        while (cur != NULL) {
            ListNode *next = cur->next;
            cur->next = pre;
            pre = cur;
            cur = next;
        }
        
        return pre;
    }
};