题解 | #直接遍历——合并两个排序的链表#

/*
struct ListNode {
	int val;
	struct ListNode *next;
	ListNode(int x) :
			val(x), next(NULL) {
	}
};*/
class Solution {
public:
    ListNode* Merge(ListNode* pHead1, ListNode* pHead2) {
        ListNode *pre = new ListNode(0); //新链表的头节点
		ListNode*p=pre;//新链表中连接下一个结点的指针
		while (pHead1 != nullptr && pHead2 != nullptr) {
			if(pHead1->val <= pHead2->val){
				p->next = pHead1;
				pHead1 = pHead1->next;
			}else {
				p->next = pHead2;
				pHead2 = pHead2->next;
			}
			p = p->next;
		}
		if(pHead1 == nullptr) p->next = pHead2; //pHead1为空，那就将剩下的pHead2连接到p后面
		else p->next = pHead1;
		return pre->next;
    }
};