Leetcode 143. Reorder List

VER0:

The simplest idea is to get the middle node, put the node after it into the stack, and start with the head and insert the Stack.top ().

Specific explanation: After getting slow, in two cases, the list length is even, the slow is the last node of the first half; the list length is odd, and the slow is the previous node of the middle node. In both cases the Reoder Slow->next is still in its original position, so mid = Slow->next->next.

72ms

1 /**2 * Definition for singly-linked list.3 * struct ListNode {4 * int val;5 * ListNode *next;6 * ListNode (int x): Val (x), Next (NULL) {}7  * };8  */9 classSolution {Ten  Public: One     voidReorderlist (listnode*head) { A         if(head = = NULL | | head->next = = NULL)return; -listnode* slow = head, * fast = head->Next; -          while(Fast->next && fast->next->next) { theslow = slow->Next; -Fast = Fast->next->Next; -         } -listnode* mid = Slow->next->Next; +Slow->next->next =NULL; -Stack<listnode*>Stk; +          while(mid) { A Stk.push (mid); atMID = Mid->Next; -         } -listnode* Resh =head; -          while(!Stk.empty ()) { -listnode* tmp = resh->Next; -Resh->next =stk.top (); inResh->next->next =tmp; - Stk.pop (); toResh =tmp; +         } -          the          *     } $};

All two of the following algorithms are 64ms

Ver1:

Put all the nodes into the vector, the variables I, J start at the ends of the vector, link.

1 classSolution {2  Public:3     voidReorderlist (listnode*head) {4Vector<listnode*>v;5listnode* tmp =head;6          while(TMP) {7 V.push_back (TMP);8TMP = tmp->Next;9         }       Ten          for(intI=0, J=v.size ()-1; i<=j; i++, j-- ) { One             if(tmp) Tmp->next =V[i]; AV[i]->next =V[j]; -TMP =V[j]; -         } the         if(tmp) Tmp->next =NULL; -     } -};

Ver2:

Get the midpoint of the list, reverse the second half of the list, and then merge the two-segment linked list.

1 classSolution {2  Public:3listnode* Reverselist (listnode*head) {4     if(Head==null | | head->next==null)returnhead;5ListNode *pre=head, *cur=pre->next, *Post;6Pre->next =NULL;7      while(cur!=NULL) {8Post = cur->Next;9Cur->next =Pre;TenPre =cur; OneCur =Post; A     } -     returnPre; - } the voidMergelist (listnode* L1, ListNode *L2) { -ListNode *cur1=l1, *cur2=l2,*post1,*Post2; -      while(cur2!=NULL) { -Post1 = cur1->Next; +Post2 = cur2->Next; -Cur1->next =Cur2; +Cur2->next =Post1; ACur1 =Post1; atCUR2 =Post2; -     } - } - voidReorderlist (listnode*head) { -     if(Head==null | | head->next==null | | head->next->next==null)return; -ListNode *slow=head,*fast=head; in      while(Fast->next!=null && fast->next->next!=NULL) { -Fast = Fast->next->Next; toslow = slow->Next; +     } -ListNode *head2 = slow->Next; theSlow->next =NULL; * mergelist (Head,reverselist (head2)); $ }Panax Notoginseng};

There is also a recursive algorithm that does not look, leaving the pit

Https://leetcode.com/discuss/93197/concise-recursive-o-solution-with-space-easy-to-understand

Leetcode 143. Reorder List