LeetCode Reorder List

最新推荐文章于 2025-09-11 19:19:55 发布

无止境的yy

最新推荐文章于 2025-09-11 19:19:55 发布

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分类专栏：一般算法文章标签： leetcode 链表

本文链接：https://blog.youkuaiyun.com/go54321/article/details/17312097

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本文介绍了一种链表重排序算法的实现方法，通过将链表后半部分反转再与前半部分交错合并来达到重排序的目的。文章详细阐述了算法步骤，并提供了两种不同版本的代码实现。

智商拙计，做了有些时间才对，抱着试一试的心态先用了个递归的暴力找最后节点的方法，超时，关键是解决如何高效的取得最后一个节点，可以先把后半截链表倒转一下，然后merge两个链表，总共节点数为偶数时merge时刚刚能够一一配对，奇数时则会多出一个，补上即可。中间分割链表的时候不小心又出了好些bug，有些事情总是要被拆穿啊~

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverse(ListNode* head) {
        if (head == NULL) {
            return NULL;
        }
        ListNode* pre = NULL;
        ListNode* cur = head;
        while (cur != NULL) {
            ListNode* tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        return pre;
    }

    void reorderList(ListNode *head) {
        if (head == NULL || head->next == NULL) {
            return;
        }
        int num = 0;
        ListNode* cur = head;
        while (cur != NULL) {
            num++;
            cur = cur->next;
        }
        // seperate list, find middle node as the head of the new list
        int ridx = num / 2;
        ListNode* rhead = NULL;
        cur = head;
        for (int i=0; true; i++) {
            if (i == ridx - 1) {
                rhead = cur->next;
                cur->next = NULL;
                break;
            }
            cur = cur->next;
        }
 
        rhead = reverse(rhead);
        // build final list
        cur = head;
        head = head->next;
    
        cur->next = rhead;
        cur = rhead;
        rhead = rhead->next;
            
        while (head != NULL && rhead != NULL) {
            cur->next = head;
            cur = head;
            head = head->next;
                
            cur->next = rhead;
            cur = rhead;
            rhead = rhead->next;
        }
        cur->next = rhead; // there must be only one node left(odd case) or NULL(even case)
    }
};

还是附上naive版本吧，too young, too simple

/**
 * 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* tail = head;
        ListNode* prev = NULL;
        while (tail->next != NULL) {
            prev = tail;
            tail = tail->next;
        }
        prev->next = NULL;
        reorderList(head->next);
        tail->next = head->next;
        head->next = tail;
    }
};