merge two sorted linked list in c

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

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

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

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

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; -

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

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

/*** 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

/** 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

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; - }

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

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

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