leetcode 23

leetcode 23合并K个排序链表（merge-k-sorted-lists）

使用堆排序构造一个优先队列可以轻松解决，一直以来了解堆排序但没有实际写过，这次大概写了一个半成品，用来做这道题足够了。

//c++版本
class Heap {
private:
	int size;
	vector ptrs;
public:
	Heap(vector list)
	{
		for(int i =0;ival < ptrs[ind]->val)
		{
			ind = left;
		}
		if (right < size&&ptrs[right]->val < ptrs[ind]->val)
			ind = right;
		if (ind != i)
		{
			temp = ptrs[i];
			ptrs[i] = ptrs[ind];
			ptrs[ind] = temp;
			minHeapfy(ind);
		}
	}

	void HeapSort()
	{
		for (int i = (size - 2) / 2; i >= 0; i--)
		{
			minHeapfy(i);
		}
	}
	
	int extractmin()
	{
		int res = ptrs[0]->val;
		if(ptrs[0]->next==NULL)
        {
            ptrs[0] = ptrs[size-1];
            size -= 1;
        }
        else
        {
            ptrs[0] = ptrs[0]->next;
        }
        minHeapfy(0);
		return res;
	}
	
    int getsize()
    {
        return size;
    }
};



/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* mergeKLists(vector& lists) {
        if(lists.empty())
            return NULL;
        Heap heap(lists);
        if(heap.getsize()==0)
            return NULL;
        ListNode* res=NULL;
        ListNode* ptr=NULL;
        int flag = 0;
        heap.HeapSort();
        while(heap.getsize()!=0)
        {
            ListNode* node = new ListNode(heap.extractmin());
            if(!flag)
            {
                res = ptr = node;
                flag = 1;
            }
            else
            {
                ptr->next = node;
                ptr = node;
            }
        }
        return res;
    }
};