Leetcode 21 合併兩個有序連結串列 學習感悟

思路：

迭代一個個往裡面加

# include<iostream>
# include<vector>
# include<string>
# include<algorithm>
# include<math.h>
# include<climits>
# include<stack>
using namespace std;
struct ListNode {
	int val;
	ListNode* next;
	ListNode() : val(0), next(nullptr) {}
	ListNode(int x) : val(x), next(nullptr) {}
	ListNode(int x, ListNode* next) : val(x), next(next) {}
};
ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
	ListNode* head = NULL;
	ListNode* p=head;
	ListNode* lp1 = l1;
	ListNode* lp2 = l2;
	while (lp1 || lp2) {//只要有一個節點存在
		if (lp1 && lp2) {
			if (lp1->val < lp2->val) {//lp1數字小
				l1 = lp1->next;
				if (!head) {//head開始為空
					lp1->next = NULL;//斷尾
					head = lp1;
					lp1 = l1;
					p = head;
				}
				else {//head 已經不為空了
					p->next = lp1;
					p = lp1;
					lp1 = l1;
					p->next = NULL;
				}
			}
			else {//lp2數字小
				l2 = lp2->next;
				if (!head) {//head開始為空
					lp2->next = NULL;//斷尾
					head = lp2;
					lp2 = l2;
					p = head;
				}
				else {//head 已經不為空了
					p->next = lp2;
					p = lp2;
					lp2 = l2;
					p->next = NULL;
				}
			}
		}
		else if (!lp1) {//lp1不存在
			if (!head) {//頭不存在
				head = l2;
				l2 = l2->next;
				p = head;
			}
			while (l2) {
				p->next = l2;
				p = p->next;
				l2 = l2->next;
			}
			lp2 = NULL;
			p->next = NULL;
		}
		else if (!lp2) {//lp2不存在
			if (!head) {//頭不存在
				head = l1;
				l1 = l1->next;
				p = head;
			}
			while (l1) {
				p->next = l1;
				p = p->next;
				l1 = l1->next;
			}
			lp1 = NULL;
			p->next = NULL;
		}
		else { ; }//不可能出現
	}
	return  head;
}
int main(void) {
	ListNode* l1=new ListNode(1, new ListNode(2, new ListNode(4)));
	ListNode* l2=new ListNode(1, new ListNode(3, new ListNode(4)));;
	ListNode* head = mergeTwoLists(l1, l2);
	system("pause");
	return  0;
}

優化：

注意：l1,l2 在mergeTwoLists函式中是拷貝，所以在函式中將l1 , l2 資訊丟失並不會影響原先的l1,l2 的連結串列資料

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
	ListNode* head = new ListNode(-1);
	ListNode* p=head;
	while (l1 && l2) {
		if (l1->val < l2->val) {//l1小
			p->next = l1;
			l1 = l1->next;
		}
		else {
			p->next = l2;
			l2 = l2->next;
		}
		p = p->next;
	}
	if (l1) 
		p->next = l1;
	if (l2)
		p->next = l2;
	p = head;
	head = head->next;
	delete(p);
	return  head;
}