题目描述

法一 递归

class Solution {
public:
    ListNode* mergeTwoLists(ListNode *l1, ListNode*l2){
		if(l1 == nullptr){
			return l2;
		} else if (l2==nullptr){
			return l1;
		} else if (l1->val<l2->val){
			l1->next = mergeTwoLists(l1->next, l2);
			return l1;
		} else {
			l2->next = mergeTwoLists(l1, l2->next);
			return l2;
		}
	}
};

法二 迭代

1.使用dumy->next来保存需要返回的头结点，注意创建新结点的语法 ListNode* dummy = new ListNode(-1)
2.判断l1,l2哪个值更小，并用 pp指针 来存储更小的结点指针，帮助dummy连接后，控制更小的l1,l2往后移动
直到，一边有nullptr，把另一边都接上

class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2){
		ListNode* dummy = new ListNode(-1);   //创建一个新的头节点 
		ListNode* cur = dummy;
		while(l1 && l2){
			ListNode* pp = (l1->val < l2->val) ? l1 : l2;
			cur->next = pp;
			cur = cur->next;
			if(pp==l1) {
				l1=l1->next; 
			} else{
				l2 = l2->next;
			} 
		}
		cur->next = (l1==nullptr) ? l2 : l1;
		ListNode* res = dummy->next;
		delete dummy;
		return res;	
	}

	//version2
	ListNode* mergeTwoLists(ListNode* l1, ListNode* l2){
            if((!l1) || (!l2)) return l1 ? l1 : l2;
            ListNode* dummy = new ListNode(-1);   //创建一个新的头节点 
            ListNode *cur=dummy, *aPtr = l1, *bPtr = l2;
            while(aPtr && bPtr){
                if(aPtr->val < bPtr->val){
                    dummy->next = aPtr;
                    aPtr = aPtr->next;
                } else {
                    dummy->next = bPtr;
                    bPtr = bPtr->next;
                }
                dummy = dummy->next;
            }
            dummy->next = aPtr ? aPtr : bPtr;
            return cur->next;	
        }
};