Merge Two Sorted Lists

Description

You are given the heads of two sorted linked lists list1 and list2.

Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.

Return the head of the merged linked list.

Example 1:

Input: list1 = [1,2,4], list2 = [1,3,4] Output: [1,1,2,3,4,4]

Example 2:

Input: list1 = [], list2 = [] Output: []

Example 3:

Input: list1 = [], list2 = [0] Output: [0]

Constraints:

• The number of nodes in both lists is in the range [0, 50].
• -100 <= Node.val <= 100
• Both list1 and list2 are sorted in non-decreasing order.

Code

Time Complexity: $O(N)$, Space Complexity: $O(1)$

Direct Pointer

C:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) {
    struct ListNode* dummy = malloc(sizeof(struct ListNode));
    struct ListNode* res = dummy;
    while(list1 && list2) {
        if(list1->val < list2->val) {
            dummy->next = list1;
            list1 = list1->next;
        } else {
            dummy->next = list2;
            list2 = list2->next;
        }
        dummy = dummy->next;
    }
    if(list1) {
        dummy->next = list1;
    } else {
        dummy->next = list2;
    }
    return res->next;
}

C++:

/**
 * 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(0);
        ListNode* curr = dummy;
        while(list1 && list2) {
            if(list1->val < list2->val) {
                curr->next = new ListNode(list1->val);
                list1 = list1->next;
            } else {
                curr->next = new ListNode(list2->val);
                list2 = list2->next;
            }
            curr = curr->next;
        }
 
        if(list1) {
            curr->next = list1;
        } else if (list2) {
            curr->next = list2;
        }
 
        return dummy->next;
    }
};

Pointer to Pointer (Indirect Pointer)

想避免組態暫時節點的空間 (即上方程式碼中的 malloc)，該怎麼做？運用上述 indirect pointer 的技巧（因為 indirect pointer 只需要改變 address，所以不需要用到 malloc）：

C:

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* mergeTwoLists(struct ListNode* list1, struct ListNode* list2) {
    struct ListNode* head;
    struct ListNode** indirect = &head;
 
    while(list1 && list2) {
        if(list1->val < list2->val) {
            *indirect = list1;
            list1 = list1->next;
        } else {
            *indirect = list2;
            list2 = list2->next;
        }
        indirect = &(*indirect)->next;
    }
    if(list1) {
       *indirect = list1;
    } else {
       *indirect = list2;
    }
    return head;
}

C++:

/**
 * 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* head = nullptr;
        ListNode** indirect = &head;
 
        while(list1 && list2) {
            if(list1->val < list2->val) {
                *indirect = list1;
                list1 = list1->next;
            } else {
                *indirect = list2;
                list2 = list2->next;
            }
 
            indirect = &((*indirect)->next);
        }
 
        if(list1) 
            *indirect = list1;
        if(list2) 
            *indirect = list2;
 
        return head;
    }
};

Source

• Merge Two Sorted Lists - LeetCode