#21 two linked lists after the merge sort

Ideas

Use three cursors: cur points to the end of the merged list, L1,L2 is used to traverse two linked lists, and the smaller elements are added to the combined list.

Little Tricks

Using redundant head nodes can be used to determine the situation in a concise manner, where a linked list, or two, is an empty list.

thereby streamlining the code.

Plain Code

classSolution { Public: ListNode* Mergetwolists (listnode* L1, listnode*L2) {ListNode* Head, *ptr; if(!l1 &&!l2)returnNULL; if(!L1 && L2)returnL2; if(L1 &&!l2)returnL1; if(L1->val > l2->val) {Head=L2; L2= l2->Next; } Else{Head=L1; L1= l1->Next; } ptr=Head;  while(L1 &&L2) {            if(L1->val <= l2->val) {ptr->next =L1; L1= l1->Next; } Else{ptr->next=L2; L2= l2->Next; } ptr= ptr->Next; }                if(L1) {ptr->next =L1; } Else{ptr->next =L2; }                returnHead; }};

Optimized code

classSolution { Public: ListNode* Mergetwolists (listnode* L1, listnode*L2) {ListNode Dummy (0); ListNode*ptr = &dummy;  while(L1 &&L2) {            if(L1->val <= l2->val) {ptr->next =L1; L1= l1->Next; } Else{ptr->next=L2; L2= l2->Next; } ptr= ptr->Next; }                if(L1) {ptr->next =L1; } Else{ptr->next =L2; }                returnDummy.next; }};

#21 two linked lists after the merge sort