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two-sorted-lists

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Merge Two Sorted Lists – LeetCode 21

Problem

Description

Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.

Answer

Original

Code

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/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if((l1 == nullptr) && (l2 == nullptr)) return nullptr;
        ListNode *dummyRoot = new ListNode(0);
        ListNode *ptr = dummyRoot;
        while((l1 != nullptr) || (l2 != nullptr)){
            if(l1 == nullptr) {ptr->next = new ListNode(l2->val); ptr = ptr->next; l2 = l2->next; continue;}
            if(l2 == nullptr) {ptr->next = new ListNode(l1->val); ptr = ptr->next; l1 = l1->next; continue;}
            if(l1->val < l2->val) {ptr->next = new ListNode(l1->val); ptr = ptr->next; l1 = l1->next;}
                else {ptr->next = new ListNode(l2->val); ptr = ptr->next; l2 = l2->next;}
        }
        ptr = dummyRoot->next;
        delete dummyRoot;
        return ptr;
    }
};

思路

要求合并两个已排序的List，没有难度，单纯的遍历拼接，只需注意DummyRoot的处理就行了。不过事后看了不少答案都倾向复用参数已有的ListNode部分。我认为这个函数不应该引用原有的l1和l2，因为可能会因此造成潜在的问题，比如资源的不正确访问和释放。时间复杂度是$O(m + n)$，m和n分别为两个List的长度，空间复杂度是$O(1)$。
耗时$9$ ms，排名$38.42\%$。

Better

思路

另外的思路是递归解，就算了。