merge two sorted linked list in c

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Merge two sorted lists

The title describes the input of two monotonically increasing lists, the output of the list of two linked lists, of course, we need to synthesize the linked list to meet monotonic rules. Idea: Non-recursiveClass Solution {public: listnode* Merge (listnode* pHead1, listnode* pHead2) { if (!phead1) return pHead2; if (!phead2) return pHead

Leetcode: Merge two Sorted Lists_leetcode

Merge two sorted linked lists and return it as a new list. To merge two sequential lists and return a new list The idea of solving problems: merging by order ... Code: public class Solution {public listnode mergetwolists (listnode L1, ListNode L2) {listnode f

Leetcode | #21 Merge Sorted Lists

Topic:Merge sorted linked lists and return it as a new list. The new list should is made by splicing together the nodes of the first of the lists.Ideas: Set two head pointers, one pointer fixed, to return the last header, a pointer to organize the node order, traverse the two linked lists, each pointing to a small node public class Mergetwosortedlis

023. Merge k Sorted Lists

Method One: Time complexity O (m1 + m2 +: m2), timed out1 /**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: Onelistnode* mergeklists (vectorlists) {Pay attention to the empty situation here [[]] A //finding the least-valued node -listnode* ptr =nullptr; -listnode* Phead =ptr; the if(lists.size () = =0)return

Leetcode–refresh–merge Sorted Lists

Iterative: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: OneListNode *mergetwolists (ListNode *l1, ListNode *L2) { A if(!L1)returnL2; - if(!L2)returnL1; -ListNode *result =NewListNode (0); theListNode *runner =result; - while(L1 | |L2) { - intTMP1 = Int_max, TMP2 =Int_max; -

Leetcode Merge Sorted Lists merging sort

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: OneListNode *mergetwolists (ListNode *l1, ListNode *L2) { A if(l1==0)returnL2; - if(l2==0)returnL1; -ListNode *start=0, *end=0; the if(l1->valval) { -Start=end=L1; -L1=l1->Next; - } + Else{ -Start=end=L2; +L2=l2->Next; A } at

Leetcode merge K sorted lists

MergeKSorted linked lists and return it as one sorted list. analyze and describe its complexity. Merge K sort lists The solution is as follows: Fetch k elements for heap sorting, fetch the smallest elements each time, and insert them into the linked list. Note that this question uses the priority queue of C ++. The und

Merge Sorted Lists

/*** Definition for singly-linked list.* struct ListNode {* int val;* ListNode *next;* ListNode (int x): Val (x), Next (NULL) {}* };*/Class Solution {Publiclistnode* mergetwolists (listnode* L1, listnode* L2) {Maintain a new list with two pointers to the linked list, and be aware that the head node is not the same as the nodes behind it.if (l1==null)return L2;if

Merge Sorted Lists

/** Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x) : Val (x), Next (NULL) {}}; */classSolution { Public: ListNode* Mergetwolists (listnode* L1, listnode*L2) { if(l1==nulll2==NULL)returnNULL; if(l1==NULL)returnL2; if(l2==NULL)returnL1; ListNode*Head; if(l1->valval) {Head=L1; L1=l1->Next; } Else{Head=L2; L2=l2->Next; } ListNode* p=Head; while(l1L2) {ListNode* k= (l1->valL1:l2; P->next

Merge Sorted Lists

1 /**2 * Definition for singly-linked list.3 * Function ListNode (val) {4 * This.val = val;5 * this.next = null;6 * }7 */8 /**9 * @param {listnode} L1Ten * @param {listnode} L2 One * @return {ListNode} A */ - varMergetwolists =function(L1, L2) { - varRET =NewListNode (0), thep =ret; - - while(L1!==NULL L2!==NULL) { - if(L1.val l2.val) { +P.next =L1; -L1 =L1.next; +}Else { AP.next = atL2 =L2.next; - } - -p =P.next; - }

Leetcode Merge Sorted Lists

translation合并两个排好序的链表,并返回这个新链表。新链表应该由这两个链表的头部拼接而成。Originaltwoandreturnitasanewnewbytheofthefirsttwo lists.Code/** * Definition for singly-linked list. * struct ListNode {* int val; * ListNode *next; * ListNode (int x ): Val (x), Next (NULL) {}}; */ class solution {public:listnode* mergetwolists (listnode* L1, listnode* L2) {if(L2 = = NULL) {returnL1; }if(L1 = = NULL) {returnL2; }if(l1->Val> l2->Val) {listnode* temp = L2; Temp->next = mergetwolists (L1

(Leetcode) Merge Sorted Lists

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: Onelistnode* mergetwolists (listnode* L1, listnode*L2) { AListNode *l3 = NULL;//Head Pointer -ListNode *p =L3; - //if (L1 = = NULL L2 = = null) the //return NULL; - //else if - //return (L1 = = NULL)? l1::l2; - if(L1 = = NULL)return

Merge two sorted lists

1 /** 2 * Definition for singly-linked list. 3 * public class ListNode { 4 * int val; 5 * ListNode next; 6 * ListNode(int x) { 7 * val = x; 8 * next = null; 9 * }10 * }11 */12 public class Solution {13 public ListNode mergeTwoLists(ListNode l1, ListNode l2) {14 if(l1==null) return l2;15 if(l2==null) return l1;16 ListNode head=l1.val Merge

Leetcode -- merge two sorted lists

I wrote it again, and the code looked much refreshed. 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 class Solution {10 public:11 ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {12 ListNode tmp(-1);13 ListNode *resHead = tmp;14 ListNode *resHead_output = resHead;15 while(l1 != NULL

[Leetcode] 21. Merge two Sorted Lists java__ merged

/**21. Merge two Sorted Lists * @param L1 * @param L2 * @return Merging two ordered lists/public ListNode mergetwolists (Li Stnode L1, ListNode L2) { listnode ret = new ListNode ( -1); ListNode L = ret; while (L1!= null L2!= null) { if (L1.val Improved: Linked list, so do not read each loop public ListN

Merge two sorted linked lists

Question: Merge two sorted linked lists and return it as a new list. The new list shoshould be made by splicing together the nodes of the first two lists. The Code is as follows: Struct ListNode {Int val;ListNode * next;ListNode (int x): val (x), next (NULL ){}}; ListNode * mergeTwoLists (ListNode * l1, ListNode * l2 )

JavaScript realizes link list insert sort and chain list merge sort _jsp programming

table node disorder. * Insert the nodes in List 2 into the list 1 and keep the list 1 orderly. * The last link in table 1 contains all the nodes, and is in order. */ Linklist Linkinsertsort (linklist head) { //current points to the node that is currently being inserted. linklist head2, Current, p, Q; if (head = = NULL) return head ; Split the

Merge Sorted Lists

1/**2* Definition for singly-linked list.3* Function ListNode (val) {4* This.val = val;5* This.next = null;6* }7*/8/**9* @param {ListNode} l110* @param {ListNode} L211* @return {ListNode}12*/13var mergetwolists =function(L1, L2) {14var ret =New ListNode (0),p =Ret1617while (L1!==Null L2!==Null) {18if (L1.val L2.val) {P.next =L1;L1 =L1.next;21}Else{P.next =L2;L2 =L2.next;24}25p =P.next;27}2829if (L1!==Null) {P.next =L1;31}3233if (L2!==null) {p.next =

21. Merge Sorted Lists

1listnode* mergetwolists (listnode* L1, listnode*L2) {2listnode*T;3t= (listnode*) malloc (sizeof(ListNode));4ListNode *T1;5t1=T;6 if(l1==nulll2==NULL)7 returnNULL;8 while(l1L2)9 {Ten if(l1->valval) One { At1->next=L1; -L1=l1->Next; - } the Else - { -t1->next=L2; -L2=l2->Next; + } -T1=t1->Next; + } A if(l1==NULL) at { -t1->next=L2; - } - if(l2==NULL)

Merge K Sorted Lists

Idea: Put the head of each list in the smallest heap, take the first heap at a time, and then take the next node out into the heap.1 classcmp{2 Public:3 BOOL operator() (listnode* l1,listnode*L2)4 {5 returnL1->val>l2->Val;6 }7 };8 classSolution {9 Public:Tenlistnode* mergeklists (vectorlists) { Onelistnode* head,*ptr1,*ptr2; APriority_queueminheap; - for(intI=0; I) - if(lists[i]!=NULL) the Minheap.push (Lis

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