Sword Point of Offer question 17: Merge two sorted lists

Topic:

Input two monotonically increasing list, output two list of linked lists, of course, we need to synthesize the linked list to meet the monotone non-reduction rules.

Idea 1: let two pointers point to two linked lists, who are small to insert the end of the current node into a new linked list

Code:

 /*struct ListNode {int val;struct ListNode *next; ListNode (int x)  :val (x),  next (NULL)  {}};*/class Solution {public:     listnode* merge (listnode* phead1, listnode* phead2)     {         if (phead1==null)          {            return pHead2;         }        else if (PHead2==NULL )         {             return pHead1;        }         //Two Hands         ListNode *newhead=NULL;         listnode *cur=null;        listnode *p1=phead1;         listnode *p2=phead2;        listnode  *temp=null;        //Note that if this is the following: There is a big loophole          //looks like Newhead's next is cur        //. But when the second number is found, cur points to the other         //newhead. There is only one node in the linked list          /*while (p1!=null&&p2!=null) {if (p1->_data<=p2->_data) {cur=p1;p1= P1->_next;} Else{cur=p2;p2=p2->_next;} if (newhead==null) {newhead=cur;} Cur->_next=null;cur=cur->_next;} */           while (p1!=null&&p2!=null)          {             if (P1->val<=p2->val)             {                 temp=p1;                 p1=p1->next;                              }            else             {                 temp=p2;                 p2=p2->next;             }            if ( Newhead==null)             {                 newhead=temp;                 cur=newhead;             }            else             {                 cur->next=temp;                 cur=cur->next;             }        }         if (p1!=null)         {             cur->next=p1;        }         else        {             cur->next=p2;         }        return newhead;    }};

idea two: by recursion, find the smallest element each time, add to the back of the new list

Code:

Listnode* merge (listnode* phead1, listnode* phead2)     {         //termination Condition         if (PHead1==NULL)         {             return pHead2;        }         else if (phead2==null)         {             return pHead1;         }        ListNode *newhead=NULL;         if (Phead1->val<=phead2->val)          {            newhead =pHead1;      &Nbsp;      newhead ->next=merge (phead1->next,phead2);         }        else         {            newhead  =phead2;            newhead ->next= Merge (Phead1,phead2->next);        }         return newhead;    }

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Sword Point of Offer question 17: Merge two sorted lists