[leetcode 143] Reorder List

最新推荐文章于 2025-12-28 22:03:12 发布

原创最新推荐文章于 2025-12-28 22:03:12 发布 · 175 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#链表 #leetcode

[LeetCode 刷题] 专栏收录该内容

158 篇文章

订阅专栏

本文介绍了一种链表重组算法，该算法将给定的单链表重新排列为特定的交错形式，即首尾元素相邻，随后是次首次尾元素相邻，以此类推。文章详细解释了如何通过双指针技巧找到链表中点，再反转后半部分链表，并交织合并前半部分和反转后的后半部分链表。

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}.

/**
 * 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 || !head->next) {
            return ;
        }
        ListNode *prev = NULL;
        ListNode *slow = head;
        ListNode *fast = head;
        while (fast && fast->next) {
            prev = slow;
            slow = slow->next;
            fast = fast->next->next;
        }
        prev->next = NULL;
        slow = reverse(slow);
        ListNode *curr = head;
        while (curr->next) {
            ListNode *tmp = curr->next;
            curr->next = slow;
            slow = slow->next;
            curr->next->next = tmp;
            curr = tmp;
        }
        curr->next = slow;
    }
    ListNode *reverse(ListNode *head) {
        if (!head || !head->next) {
            return head;
        }
        ListNode *prev = head;
        for (ListNode *cur = head->next, *next = cur->next; cur; prev = cur, cur = next, next = cur?cur->next:NULL) {
            cur->next = prev;
        }
        head->next = NULL;
        return prev;
    }
};