题解 | #合并两个排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
#include <cstddef>
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param pHead1 ListNode类 
     * @param pHead2 ListNode类 
     * @return ListNode类
     */
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        if(pHead1==NULL||pHead2==NULL){
            return pHead1;
        }
        ListNode *head_node = new ListNode(-1);
        ListNode *begin = head_node;
        
        while (pHead1!=NULL||pHead2!=NULL) {
            if(pHead1==NULL){
                begin->next = pHead2;
                break;
            }
            if(pHead2==NULL){
                begin->next = pHead1;
                break;
            }
            if(pHead1->val<=pHead2->val){
                    ListNode *tmp1 = pHead1->next;
                    begin->next = pHead1;
                    pHead1 = tmp1;
                }
            else{
                    ListNode *tmp2 = pHead2->next;
                    begin->next = pHead2;
                    pHead2 = tmp2;
                }            
            begin = begin->next;
     
        }
        head_node = head_node->next;
        return head_node;
    }
};