题目描述:
将两个有序链表合并为一个新的有序链表并返回。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:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-two-sorted-lists
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
解答:
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* mergeTwoLists(struct ListNode* l1, struct ListNode* l2)
{
if (NULL == l1)
{
return l2;
}
if (NULL == l2)
{
return l1;
}
struct ListNode* p1 = l1;
struct ListNode* p2 = l2;
int len = 0;
int i = 0;
int k = 0;
int *Num = NULL;
while(p1)
{
len++;
p1 = p1->next;
}
while(p2)
{
len++;
p2 = p2->next;
}
Num = (int*)malloc(sizeof(int)*len);
p1 = l1;
p2 = l2;
while(p1)
{
Num[i]=p1->val;
p1=p1->next;
i++;
}
while(p2)
{
Num[i]=p2->val;
p2=p2->next;
i++;
}
//从大到小排序
int comp(const void*a,const void*b)
{
return *(int*)a-*(int*)b;
}
qsort(Num,len,sizeof(int),comp);
p1 = l1;
p2 = l2;
//将数组元素依次添加到链表中
struct ListNode *lists = NULL;
lists = malloc(sizeof(struct ListNode));
lists->val=Num[0];
lists->next = NULL;
struct ListNode* temp = lists;
struct ListNode* head = lists;
for (i = 1; i<len; i++)
{
struct ListNode *lists = NULL;
lists=malloc(sizeof(struct ListNode));
lists->val=Num[i];
temp->next = lists;
lists->next=NULL;
temp=lists;
}
free(Num);
return head;
}
运行结果: