leetcode

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}.

个人思路：

1，找到链表的中点，断开分为两部分，对链表的后半部分进行反转，然后再合并这两部分链表即可

代码：

#include <stddef.h>

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;
        }

        int length = 0;
        ListNode *current = head;

        while (current)
        {
            current = current->next;
            ++length;
        }

        ListNode *firstHead = head;
        int firstLength = length / 2 + length % 2;
        ListNode *secondHead = NULL;
        current = firstHead;

        for (int i = 0; i < firstLength; ++i)
        {
            current = current->next;
        }
        secondHead = current;

        //对second链表进行反转
        ListNode dummy(-1);
        dummy.next = secondHead;
        ListNode *pre = secondHead;
        current = secondHead->next;

        while (current)
        {
            pre->next = current->next;
            current->next = dummy.next;
            dummy.next = current;
            current = pre->next;
        }

        //合并first链表和second链表
        ListNode *firstCur = firstHead;
        ListNode *secondCur = dummy.next;
        ListNode *temp = NULL;

        while (firstCur && secondCur)
        {
            temp = secondCur->next;
            secondCur->next = firstCur->next;
            firstCur->next = secondCur;

            firstCur = secondCur->next;
            secondCur = temp;
        }

        if (firstCur)
        {
            firstCur->next = NULL;
        }
    }
};