Merge K Sorted Lists

本文介绍了一种利用堆实现的高效算法,用于合并多个已排序的链表,并详细解释了如何通过胜者树或败者树的方法简化问题,最终得到一个有序的单一链表。

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题目描述:

        有一系列已排好序的链表,将其合并成有序的一张链表;

 

第一反应是运用胜者树或败者树可以很方便的求解;

因为STL提供了堆操作的函数,所以这里借助堆来求解;

 

思路是:首先将k个链表的第一个节点集合,建堆,然后取出堆顶的节点,链接到结果链表上,然后将该节点的下一个节点入堆,直到所有链表都已经完成;

下面的Node结构体好像根本无必要,刚开始没注意是链表(而链表可以很方便的获取下一个节点,如果是数组的话还得记录它来自哪个数组);

有亮点需要注意:

1.当一条链表结束的时候,添加值为INF的新Node加入堆,可以方便地避免NULL的判断;

2.make_heap()默认是大顶堆,所以要显示指定使用greater<node>(),来获取小顶堆;

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 
#define LN ListNode*
#define HEAP heap.begin(),heap.end()
#define PB push_back
#define INF 1000000

struct node
{
	int val;
	LN from;
	
	node(ListNode* n)
	{
		if ( n==NULL )
		{
			val=INF;
		}
		else
		{
			val=n->val;
		}			
		from=n;
	}
	bool operator<(const node& other)const
	{
		return val<other.val;
	}
	bool operator>(const node& other)const
	{
		return val>other.val;
	}
};
    
class Solution {
public:
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        if (lists.empty()) return NULL;
	int n= lists.size();
	vector<node> heap;
	heap.reserve(n);
	for( int i=0;i<n;i++)
		heap.PB(node(lists[i]));
	make_heap(HEAP,greater<node>());
	LN head= new ListNode(0);
	LN pL = head;
	pop_heap(HEAP,greater<node>());
	node small=heap.back();
	heap.pop_back();
	while(small.val!=INF)
	{
		LN next=small.from->next;
		pL->next=small.from;
		small.from->next=NULL;
		pL=pL->next;

		heap.PB(node(next));
		push_heap(HEAP,greater<node>());
		pop_heap(HEAP,greater<node>());
		small=heap.back();
		heap.pop_back();
	}

	LN ret=head->next;
	delete head;
	return ret;    }
};

To merge k sorted linked lists, one approach is to repeatedly merge two of the linked lists until all k lists have been merged into one. We can use a priority queue to keep track of the minimum element across all k linked lists at any given time. Here's the code to implement this idea: ``` struct ListNode { int val; ListNode* next; ListNode(int x) : val(x), next(NULL) {} }; // Custom comparator for the priority queue struct CompareNode { bool operator()(const ListNode* node1, const ListNode* node2) const { return node1->val > node2->val; } }; ListNode* mergeKLists(vector<ListNode*>& lists) { priority_queue<ListNode*, vector<ListNode*>, CompareNode> pq; for (ListNode* list : lists) { if (list) { pq.push(list); } } ListNode* dummy = new ListNode(-1); ListNode* curr = dummy; while (!pq.empty()) { ListNode* node = pq.top(); pq.pop(); curr->next = node; curr = curr->next; if (node->next) { pq.push(node->next); } } return dummy->next; } ``` We start by initializing a priority queue with all the head nodes of the k linked lists. We use a custom comparator that compares the values of two nodes and returns true if the first node's value is less than the second node's value. We then create a dummy node to serve as the head of the merged linked list, and a current node to keep track of the last node in the merged linked list. We repeatedly pop the minimum node from the priority queue and append it to the merged linked list. If the popped node has a next node, we push it onto the priority queue. Once the priority queue is empty, we return the head of the merged linked list. Note that this implementation has a time complexity of O(n log k), where n is the total number of nodes across all k linked lists, and a space complexity of O(k).
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