LeetCode 23： Merge K Sorted Lists

标签：leetcode   merge k sorted lists   

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

本题在上一题(LeetCode 21: Merge Two Sorted Lists)基础上再加深了一步，链表个数从两个改为K个。

此题有如下几种解法：

1、最简单的方法莫过于每次都从K个链表头中找到最小的那个元素，加入结果链表中。基于此，有人通过最小堆来简化最小元素的比较。

 struct CompareListNode{
	bool operator()(const ListNode*p, const ListNode* q)const{
		return p->val > q->val;
	}
};
class Solution {
public:
    ListNode* mergeKLists(vector<ListNode*>& lists) {
       ListNode head(-1);
	vector<ListNode*> vec_helper;
	//除去空链表
	for (int i=0; i<lists.size(); i++)
	{
		if (lists[i]!=NULL)
			vec_helper.push_back(lists[i]);
	}
	//建小堆
	std::make_heap(vec_helper.begin(), vec_helper.end(), CompareListNode());
	ListNode* pTail = &head;
	while (vec_helper.size() > 0)
	{
	    //取最小的元素对应的节点
		ListNode *pNode = vec_helper.front();
		pTail->next= pNode;
		pTail = pTail->next;
		std::pop_heap(vec_helper.begin(), vec_helper.end(),  CompareListNode());
		vec_helper.pop_back();
	    //将该节点之后一个节点加堆中
		if (pNode->next)
		{
			vec_helper.push_back(pNode->next);
			std::push_heap(vec_helper.begin(), vec_helper.end(),  CompareListNode());
		}
	}
	pTail->next = NULL;
	return head.next;
    }
};

同样还有基于优先队列的算法：

	ListNode* mergeKLists3(vector<ListNode*>& lists) {
		int length = lists.size();
		ListNode head(-1);
		ListNode* pTail = &head;
		priority_queue<ListNode*, vector<ListNode*>, CompareListNode> list_queue;
		for (int i=0; i<length; i++)
		{
			if (lists[i])
				list_queue.push(lists[i]);
		}
		while (list_queue.size() >0)
		{
			ListNode* ptmpNode = list_queue.top();
			list_queue.pop();
			pTail->next = ptmpNode;
			pTail = pTail->next;
			if (ptmpNode->next)
				list_queue.push(ptmpNode->next);
		}
		pTail->next = NULL;
		return head.next;
	}

	ListNode* mergeKLists2(vector<ListNode*>& lists) {
		int length = lists.size();
		if (length == 0)
			return NULL;
		while (lists.size() >1)
		{
			ListNode* p1 = lists.front ();
			lists.erase(lists.begin());
			ListNode* p2 = lists.front();
			lists.erase(lists.begin());
                        lists.push_back(mergeTwoLists(p1, p2));
		}
		return lists[0];
	}

方法3比较简洁

复用上题两个链表的合并代码，方法3的实现代码如下：

    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int length = lists.size();
        if (length == 0) {
            return NULL;
        } else if(length == 1)
            return lists[0];
        else if(length == 2) {
            return mergeTwoLists(lists[0], lists[1]);
        }
        vector<ListNode*> leftHalf(lists.begin(), lists.begin()+length/2);
        vector<ListNode*> rightHalf(lists.begin()+length/2, lists.end());
        return mergeTwoLists(mergeKLists(leftHalf), mergeKLists(rightHalf));
    }

LeetCode 23： Merge K Sorted Lists

标签：leetcode   merge k sorted lists   

原文地址：http://blog.csdn.net/sunao2002002/article/details/46283091