Leetcode Reorder List

实现链表如下所示：

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

一开始想到一个n方的，就是每次找最后一个回调到相应位置，然后倒数第二个的next置为NULL，依次类推。果然超时。

/**

 * 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 *cur, *last, *next_cur, *pre_last;

        last = head -> next;

        pre_last = head;

        while(last -> next)

        {

            pre_last = last;

            last = last -> next;

        }

        cur = head;

        next_cur = head -> next;



        while(cur -> next || cur -> next != last)

        {

            cur -> next = last;

            pre_last -> next = NULL;

            cur = next_cur;

            next_cur = cur -> next;

            last = cur -> next;

            pre_last = cur;

            while(last && last -> next)

            {

                pre_last = last;

                last = last -> next;

            }

        }

        return ;

    }

};

View Code

也可以用map<int, ListNode*>来做吧应该，但用了其他空间了。下面是比较直接的没有用多余空间的解法。

 

分成两半，后面一半反转，然后合并前后两半。

需要注意的是，前面一半的最后一个记得赋值NULL，还有如果是奇数，那么前面一半应该多一个。自己随便举个1234和12345的例子画一下就知道怎么合并怎么分了。

/**

 * 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 || !head -> next) return ;

        ListNode *slow = head, *quick = head -> next;

        while(quick && quick -> next) // 分成两部分

        {

            slow = slow -> next;

            quick = quick -> next;

            if (!quick)

                break;

            quick = quick -> next;

        }

        quick = slow -> next;

        ListNode *last = quick -> next;

        while(last) // 反转后半部

        {

            quick -> next = last -> next;

            last -> next = slow -> next;

            slow -> next = last;

            last = quick -> next;

        }

        quick = slow -> next;

        slow -> next = NULL;



        ListNode *next_cur = head -> next, *next_quick, *cur = head;

        while(cur && quick) // 合并两部分

        {

            cur -> next = quick;

            next_quick = quick -> next;

            quick -> next = next_cur;

            quick = next_quick;

            cur = next_cur;

            if (cur)

                next_cur = cur -> next;

        }

    }   

};