[LeetCode]Merge k Sorted Lists

最新推荐文章于 2024-04-03 21:24:25 发布

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本文介绍了一种使用优先队列合并K个已排序链表的方法。通过定义自定义节点结构和比较器，该方法实现了O(nlogk)的时间复杂度，其中n为所有链表中节点总数，k为链表数量。文章提供了两种实现方式，详细展示了如何利用优先队列高效地合并多个排序链表。

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

class Solution {
//O(nlogk)
//using priority_queue to choose current minimum node
//node: keep priority_queue free of NULL pointer
public:
	struct PQNode
	{
		int index;
		ListNode* ptr;
		PQNode(int _index = 0, ListNode* _ptr = NULL):index(_index),ptr(_ptr){}
	};
	struct Comp
	{
		bool operator()(const PQNode& lhs, const PQNode& rhs) const
		{
			return lhs.ptr->val > rhs.ptr->val;
		}
	};
	ListNode *mergeKLists(vector<ListNode *> &lists) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		priority_queue<PQNode, vector<PQNode>, Comp> pq;
		ListNode* head = NULL;
		ListNode* cur = NULL;
		do
		{
			for(int i = 0; i < lists.size(); ++i)
			{
				if(lists[i] != NULL)
				{
					pq.push(PQNode(i, lists[i]));
					lists[i] = lists[i]->next;
				}
			}
			if(pq.empty()) break;
			if(head == NULL)
				head = pq.top().ptr;
			else 
				cur->next = pq.top().ptr;
			cur = pq.top().ptr;
			int curIdx = pq.top().index;
			pq.pop();
			if(lists[curIdx] != NULL)
			{
				pq.push(PQNode(curIdx, lists[curIdx]));
				lists[curIdx] = lists[curIdx]->next;
			}
		}while(true);
		if(cur != NULL) cur->next = NULL;
		return head;
	}
};

second time

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    struct Node
    {
        ListNode* pNode;
        int listIdx;
        Node(ListNode* _pNode = NULL, int _listIdx = 0):pNode(_pNode), listIdx(_listIdx){};
        bool operator < (const Node& orh) const {return pNode->val > orh.pNode->val;};
    };
    ListNode *mergeKLists(vector<ListNode *> &lists) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        priority_queue<Node> pq;
        ListNode dummy(-1);
        ListNode* prev = &dummy;
        for(int i = 0; i < lists.size(); ++i)
            if(lists[i] != NULL) pq.push(Node(lists[i], i)), lists[i] = lists[i]->next;
            
        while(!pq.empty())
        {
            prev->next = pq.top().pNode;//top() return the first element in pq
            prev = prev->next;
            int listIdx = pq.top().listIdx;
            pq.pop();
            if(lists[listIdx] != NULL) 
            {
                pq.push(Node(lists[listIdx], listIdx));
                lists[listIdx] = lists[listIdx]->next;
            }
        }
        
        prev->next = NULL;
        return dummy.next;
    }
};