leetcode Reorder List

最新推荐文章于 2020-07-18 13:27:31 发布

原创最新推荐文章于 2020-07-18 13:27:31 发布 · 374 阅读

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CC 4.0 BY-SA版权

本文介绍了一种重新排列单链表元素的方法，通过倒序中心节点之后的部分，实现链表元素的重新排列。

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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}.

#include <iostream>
using namespace std;
/**
 * Definition for singly-linked list.*/
  struct ListNode {
      int val;
      ListNode *next;
      ListNode(int x) : val(x), next(NULL) {}
  };
  
  void print(ListNode * head)
  {
      while(head != NULL)
      {
          cout << head->val << " ";
          head = head->next;
      }
      cout << endl;
  }

  class Solution {
  public:
      //先将中心节点以后的节点倒序然后进行插入
      //如 L1 L2 L3 L4 L5 
      // L1 L2 L3 L5 L4
      // L1 L5 L2 L4 L3 
      void reorderList(ListNode *head) {
          if(head == NULL || head->next == NULL || head->next->next == NULL)
              return;
          ListNode *midIterator = head;
          ListNode *tailIterator = head->next;
          while(tailIterator != NULL && tailIterator->next != NULL)
          {
              midIterator = midIterator->next;
              tailIterator = tailIterator->next->next;
          }
        
          ListNode * rightiterator = midIterator->next->next;
          midIterator->next->next = NULL;
          ListNode * tempiterator = NULL;
       
          while(rightiterator != NULL)
          {
              tempiterator = rightiterator;
              rightiterator = rightiterator->next;
              tempiterator->next = midIterator->next;
              midIterator->next = tempiterator;
          }

          ListNode * leftiterator = head;
          rightiterator = midIterator->next;
          while (1)
          {
              tempiterator = rightiterator;
              rightiterator = rightiterator->next;
              tempiterator->next = leftiterator->next;
              leftiterator->next = tempiterator;
              leftiterator = leftiterator->next->next;
              if(leftiterator == midIterator)
                  break;
          }
          midIterator->next = rightiterator;
          if(rightiterator != NULL)
             rightiterator->next = NULL;
      }
  };

int main()
{
	ListNode a(1);
    ListNode b(2);
    ListNode c(3);
    ListNode d(4);
    ListNode e(5);
    a.next = &b;
    b.next = &c;
    c.next = &d;
    d.next = &e;
    e.next = NULL;
    ListNode *head = &a;
    Solution s;
    s.reorderList(head);
    print(head);
    return 0;
}