leetcode腾讯精选50-023

最新推荐文章于 2019-06-13 19:44:02 发布

goodyang1

最新推荐文章于 2019-06-13 19:44:02 发布

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CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/goodyang1/article/details/90577989

博客给出了LeetCode上合并K个有序链表题目的链接，同时展示了合并两个链表的代码以及合并K个链表的类和函数代码，主要围绕链表合并的编程实现。

题目：
https://leetcode-cn.com/problems/merge-k-sorted-lists/
代码：
/**

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

ListNode(int x) : val(x), next(NULL) {}

};
*/

ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {

ListNode *pl1 = l1;
ListNode *pl2 = l2;

ListNode head(0);
ListNode *phead = &head;

while (pl1 != NULL&&pl2 != NULL)
{
	if (pl1->val < pl2->val)
	{
		phead->next = pl1;
		phead = phead->next;
		pl1 = pl1->next;
	}
	else
	{
		phead->next = pl2;
		phead = phead->next;
		pl2 = pl2->next;
	}

}

if (pl1 == NULL)
{
	phead->next = pl2;
}

if (pl2 == NULL)
{
	phead->next = pl1;
}

return head.next;

}

class Solution {
public:
ListNode* mergeKLists(vector<ListNode*>& lists) {

if (lists.empty())
	return NULL;
int size = lists.size();

if (size == 1)
{
	return lists[0];
}

if (size == 2)
{
	return mergeTwoLists(lists[0], lists[1]);
}

int mid = size / 2;

vector<ListNode *> sublist0;
vector<ListNode *> sublist1;

for (int i = 0; i < mid; i++)
{
	sublist0.push_back(lists[i]);
}

for (int i = mid ; i < size; i++)
{
	sublist1.push_back(lists[i]);
}

ListNode * l0 = mergeKLists(sublist0);
ListNode *l1 = mergeKLists(sublist1);

return mergeTwoLists(l0, l1);

}
};

结果：
在这里插入图片描述