Leetcode-merge two Sorted lists Fusion two ordered single linked list

First two single-linked lists are ordered

When merging two single-linked lists, it can be cumbersome to think about adding or subtracting a sequence.

It is important to open a serial header to store it, not necessarily to open up memory, mainly a concept

In fact, the merge of method sense and merge algorithm is a concept

/** * 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 (L1 = = NULL) {        return L2;        } else if (L2 = = NULL) {return L1;        } ListNode *head = NULL;            if (L1->val < l2->val) {head = L1;        L1 = l1->next;            } else {head = L2;        L2 = l2->next;        } ListNode *phead = head;                 while (L1 && L2) {if (L1->val < l2->val) {head->next = L1;                Head = head->next;            L1 = l1->next;                } else {head->next = L2;                Head = head->next;            L2 = l2->next;        }} if (L1){head->next = L1;        } if (L2) {head->next = L2;            } return phead; }};

Provide a solution to the recursion on the offer of the sword

/** * Definition for singly-linked list. * struct ListNode {*     int val; *     ListNode *next; *     listnode (int x): Val (x), Next (NULL) {} *}; */class Soluti On {public:    listnode *mergetwolists (ListNode *l1, ListNode *l2) {        if (L1 = = NULL)        {            return l2;        }        else if (L2 = = NULL)        {            return L1;        }        ListNode *head = NULL;        if (L1->val < L2->val)        {            head = L1;            Head->next = Mergetwolists (L1->NEXT,L2);        }        else        {            head = L2;            Head->next = Mergetwolists (L1,l2->next);        }        return head;    };

Leetcode-merge Two Sorted lists merge two ordered single-linked lists