[Leetcode] Merge two sorted lists

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本文介绍了一种将两个已排序的链表合并为一个新排序链表的方法。通过比较两个链表节点的值，选择较小的节点加入到新链表中，最终形成完整的有序链表。文章提供了一个C++实现示例。

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Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *vHead = new ListNode(-1);//virtualHead;
        ListNode *tmp = vHead;
        ListNode *tmp_A = l1;
        ListNode *tmp_B = l2;
        
        while(tmp_A && tmp_B)
        {
            if(tmp_A->val<=tmp_B->val)
            {                
                tmp->next = tmp_A;
                tmp_A = tmp_A->next;
            }
            else
            {
                tmp->next = tmp_B;
                tmp_B = tmp_B->next;
            }
            tmp = tmp->next;
        }        
        
        while(tmp_A)
        {
            tmp->next = tmp_A;
            tmp_A = tmp_A->next;
            tmp = tmp->next;
        }
        
        while(tmp_B)
        {
            tmp->next = tmp_B;
            tmp_B = tmp_B->next;
            tmp = tmp->next;
        }
        
        if(!vHead->next) return NULL;
        else return vHead->next;
    } 
    
};