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) {
        //如果一个为空的话直接返回另一个
        if(pHead1==NULL) return pHead2;
        if(pHead2==NULL) return pHead1;
        // 创建新的链表表头
        ListNode *head = new ListNode(0);
        ListNode *res = head;
        while(pHead1!=NULL && pHead2!=NULL) {
            if(pHead1->val <= pHead2->val) {
                res->next = pHead1;
                //移动取值的指针
                pHead1 = pHead1->next;
            } else {
                res->next = pHead2;
                pHead2 = pHead2->next;
            }
            //指针向后移
            res = res->next;
        }
        //哪个链表还有剩余直接连再后面
        if(pHead1!=NULL) {
            res->next = pHead1;
        } else {
            res->next = pHead2;
        }
        return head->next;
    }
};