将两个升序链表合并为一个新的 升序 链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例 1：
输入：l1 = [1,2,4], l2 = [1,3,4]
输出：[1,1,2,3,4,4]
示例 2：
输入：l1 = [], l2 = []
输出：[]
示例 3：
输入：l1 = [], l2 = [0]
输出：[0]
提示：
- 两个链表的节点数目范围是
[0, 50] -100 <= Node.val <= 100l1和l2均按 非递减顺序 排列
题解1
/**
* Definition for singly-linked list.
* 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) {}
* };
*/
class Solution {
public:
ListNode* mergeTwoLists(ListNode* list1, ListNode* list2) {
ListNode* dummy = new ListNode(-1);
auto head = dummy;
while (list1 != nullptr && list2 != nullptr) {
if (list1->val < list2->val) {
head->next = list1;
list1 = list1->next;
} else {
head->next = list2;
list2 = list2->next;
}
head = head->next;
}
if (list1 != nullptr) {
head->next = list1;
}
if (list2 != nullptr) {
head->next = list2;
}
return dummy->next;
}
};
只要设置一个dummy节点，剩下的就是简单的体力活了
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