【朱迪的LeetCode刷题笔记】21. Merge Two Sorted Lists #Easy #C

Judie

21. Merge Two Sorted Lists #Easy


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

 
Example 1:

Input: l1 = [1,2,4], l2 = [1,3,4]
Output: [1,1,2,3,4,4]


Example 2:

Input: l1 = [], l2 = []
Output: []


Example 3:

Input: l1 = [], l2 = [0]
Output: [0]
 

Constraints:

The number of nodes in both lists is in the range [0, 50].
-100 <= Node.val <= 100
Both l1 and l2 are sorted in non-decreasing order.

C

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2) {
    if (l1 == NULL) {
        return l2;
    }
    if (l2 == NULL) {
        return l1;
    }
    struct ListNode* returnNode = malloc(sizeof(struct ListNode));
    if (l2->val < l1->val) {
        returnNode->val = l2->val;
        l2 = l2->next;
        returnNode->next = NULL;
    } else {
        returnNode->val = l1->val;
        l1 = l1->next;
        returnNode->next = NULL;
    }
    struct ListNode* curNode = returnNode;
    while (1) {
        if (l1 == NULL) {
            curNode->next = l2;
            break;
        } else if (l2 == NULL) {
            curNode->next = l1;
            break;
        } else {
            struct ListNode *newNode = malloc(sizeof(struct ListNode));
            if (l2->val < l1->val) {
                newNode->val = l2->val;
                l2 = l2->next;
                newNode->next = NULL;
                curNode->next = newNode;
                curNode = newNode;
            } else {
                newNode->val = l1->val;
                l1 = l1->next;
                newNode->next = NULL;
                curNode->next = newNode;
                curNode = newNode;
            }
        }
    }
    return returnNode;
}

20分钟左右写出来的吧