leetcode 143. Reorder List 双指针-蒲公英云

leetcode 143. Reorder List 双指针

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}.

最好的方法是使用双指针来得到中间结点，然后使用栈和队列来做。

要是直接遍历然后使用反序列表来做，也可以，不过非常烦。

代码如下：

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
/*class ListNode 
{
     int val;
     ListNode next;
     ListNode(int x) { val = x; }
}*/
public class Solution 
{
    public void reorderList(ListNode head) 
    {
        if(head==null || head.next==null)
            return ;
        //使用快慢指针把链表分为两个
        ListNode slow=head;
        ListNode fast=head;
        while(fast.next!=null && fast.next.next!=null)
        {
            fast=fast.next.next;
            slow=slow.next;
        }
        ListNode head1=head , head2=slow.next;
        slow.next=null;
        Queue<ListNode> myQueue=new LinkedList<>();
        ListNode cur=head1;
        while(cur!=null)
        {
            myQueue.add(cur);
            cur=cur.next;
        }
        Stack<ListNode> myStack=new Stack<>();
        cur=head2;
        while(cur!=null)
        {
            myStack.add(cur);
            cur=cur.next;
        }
        ListNode fin=new ListNode(-1);
        cur=fin;
        while(myQueue.isEmpty()==false && myStack.isEmpty()==false)
        {
            cur.next=myQueue.poll();
            cur=cur.next;
            cur.next=myStack.pop();
            cur=cur.next;
        }
        while(myQueue.isEmpty()==false)
        {
            cur.next=myQueue.poll();
            cur=cur.next;
        }
        while(myStack.isEmpty()==false)
        {
            cur.next=myStack.pop();
            cur=cur.next;
        }
        //这是最后一步，一定要设置
        cur.next=null;
        head=fin.next;
    }
    public static void main(String[] args) {
        ListNode one=new ListNode(1);
        one.next=new ListNode(2);
        Solution solution=new Solution();
        solution.reorderList(one);
    }
}

下面是C++的做法，就是使用双指针来解决这个问题

代码如下：

#include <iostream>
#include <stack>
#include <queue>
using namespace std;
/*
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* fast = head;
        ListNode* slow = head;
        while (fast->next != NULL && fast->next->next != NULL)
        {
            fast = fast->next->next;
            slow = slow->next;
        }
        ListNode* h1 = head;
        ListNode* h2 = slow->next;
        slow->next = NULL;
        queue<ListNode*> que;
        ListNode* i = h1;
        while (i != NULL)
        {
            que.push(i);
            i = i->next;
        }
        stack<ListNode*> skt;
        i = h2;
        while (i != NULL)
        {
            skt.push(i);
            i = i->next;
        }
        ListNode* fin = new ListNode(-1);
        i = fin;
        while (!que.empty() && !skt.empty())
        {
            i->next = que.front();
            que.pop();
            i = i->next;
            i->next = skt.top();
            skt.pop();
            i = i->next;
        }
        while (!que.empty())
        {
            i->next = que.front();
            que.pop();
            i = i->next;
        }
        while (!skt.empty())
        {
            i->next = skt.top();
            skt.pop();
            i = i->next;
        }
        i->next = NULL;
        head = fin->next;
    }
};