LeetCode143：Reorder List

题目：

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

解题思路：

1，先利用快慢指针找到链表中间节点

2，将链表后半部分进行反转

3，将链表前半部分与反转后的后半部分进行合并

实现代码：

#include <iostream>

using namespace std;



/*

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}.

*/

struct ListNode {

     int val;

     ListNode *next;

     ListNode(int x) : val(x), next(NULL) {}

};

void addNode(ListNode* &head, int val)

{

    ListNode *node = new ListNode(val);

    if(head == NULL)

    {

        head = node;

    }

    else

    {

        node->next = head;

        head = node;

    }

}

void printList(ListNode *head)

{

    while(head)

    {

        cout<<head->val<<" ";

        head = head->next;

    }

}

class Solution {

public:

    void reorderList(ListNode *head) {

        if(head == NULL || head->next == NULL)

            return ;

        ListNode *quick = head;

        ListNode *slow = head;

        while(quick->next &&quick->next->next)//采用快慢指针查找链表中间节点，快指针走两步，慢指针走一步 

        {

            quick = quick->next->next;

            slow = slow->next;

        }

        quick = slow;

        slow = slow->next;

        quick->next = NULL;

        reverseList(slow);//将后半部分进行反转 

        

        quick = head;

        while(quick && slow)//将前半部分与反转后的后半部分进行合并 

        {

            ListNode *t = slow->next;

            slow->next = quick->next;

            quick->next = slow;

            slow = t;

            quick = quick->next->next;

        }

                       

    }

    void reverseList(ListNode* &head)//采用头插法进行链表反转 

    {

        if(head == NULL || head->next == NULL)

            return ;

        ListNode *p = head->next;

        head->next = NULL;

        while(p)

        {

            ListNode *t = p->next;

            p->next = head;

            head = p;

            p = t;

        }

    }

};

int main(void)

{

    ListNode *head = new ListNode(6);

    addNode(head, 5);

    addNode(head, 4);

    addNode(head, 3);

    addNode(head, 2);

    addNode(head, 1);

    addNode(head, 0);

    printList(head);

    cout<<endl;

    

    Solution solution;

    solution.reorderList(head);

    printList(head);

    

    return 0;

}