Leetcode 92 Reverse Linked List ii

標籤：

/**
* ID: 92
* Name: Reverse Linked List
* Data Structure: Linked List
* Time Complexity:
* Space Complexity:
* Tag: LinkList
* Difficult: Medium

* Problem:
* Reverse a linked list from position m to n. Do it in-place and in one-pass.
* Given 1->2->3->4->5->NULL, m = 2 and n = 4.
* return 1->4->3->2->5->NULL.
* Given m, n satisfy the following condition
* 1 ≤ m ≤ n ≤ length of list

 

思路一：

發現用vector來處理linklist會方便了很多，然後思路一就是用vector來解決

思路二： 

從m到n的node反轉，使用如下簡單的方法。 

ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversedListNode then = start.next;// a pointer to a node that will be reversed    // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3// dummy-> 1 -> 2 -> 3 -> 4 -> 5for(inti=0; i<n-m; i++){     start.next = then.next;     then.next = pre.next;     pre.next = then;     then = start.next;} 1-2-3-4  pre = 1, start = 2, then = 3把then 插入到pre的後面，pre不動，then始終在放到start的後面.start.next = then.next //把then先拿出來then.next = pre.next //把then在新位置上前後相連pre.next = then. // then的前面then = start.next // then  重新放到start的後面  思路一：

 1 class Solution { 2 public: 3     ListNode *reverseBetween(ListNode *head, int m, int n) { 4         if(m==n || head == NULL) 5             return head;  6         std::vector<ListNode *> v; 7         ListNode *p = head;  8         while(p) 9         {10             v.push_back(p);11             p = p->next; 12         }13         int len = (n-m)/2+1;14         for(int i=0; i< len;i++)15         {16             ListNode * temp = v[i+m-1];17             v[i+m-1] = v[n-i-1];18             v[n-i-1] = temp; 19             20         }21         for(int i=0;i<v.size()-1;i++)22             v[i]->next = v[i+1];23         v[v.size()-1] ->next = NULL;24         return v[0];25     }26 };

 

思路二：

 1 class Solution { 2 public: 3     ListNode *reverseBetween(ListNode *head, int m, int n) { 4         //make a head by myself 5         ListNode *newHead=new ListNode(0); 6         newHead->next=head; 7         ListNode *p0=newHead; 8         ListNode *p1=nullptr,*p2=nullptr; 9         int i;10         //find position of p011         for(i=1;i<m;i++)12             p0=p0->next;13         p1=p0->next;14         //reverse one by one15         while(i<n)16         {17             p2=p1->next;18             p1->next=p2->next;19             p2->next=p0->next;20             p0->next=p2;21             i++;22         }23 24         // p is the true head25 26         ListNode *p=newHead->next;27         delete newHead;28         return p;29     }30 };

 

 

Leetcode 92 Reverse Linked List ii