#23 Merge k Sorted Lists

題目鏈接：https://leetcode.com/problems/merge-k-sorted-lists/

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

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
void percolateUp(struct ListNode** lists, int i) {	//前i個元素已有堆序性，上濾第i+1個元素
	struct ListNode* tmp = lists[i];
	while (i > 0 && lists[(i - 1) / 2]->val > tmp->val) {	//將元素到根節點路徑上所有比該元素大的節點都下移
		lists[i] = lists[(i - 1) / 2];
		i = (i - 1) / 2;
	}
	lists[i] = tmp;
}
int buildHeap(struct ListNode** lists, int k) {
	int i, size = 0;
	for (i = 0; i < k; ++i)		//依次將非空元素插入堆中
		if (lists[i]) {
			lists[size] = lists[i];
			percolateUp(lists, size);
			++size;
		}
	return size;
}
//只有堆頂元素不滿堆序性，使其下濾；即不斷將較小兒子移到空穴直到較小兒子也不小於堆頂元素
void percolateDown(struct ListNode** lists, int size) {
	struct ListNode* tmp = lists[0];
	int i = 0, smallChild = 2 * i + 1;	//先記錄左兒子爲較小兒子，如果右兒子更小，更新
	if (smallChild + 1 < size && lists[smallChild]->val > lists[smallChild + 1]->val)
		++smallChild;
	while (smallChild < size && lists[smallChild]->val < tmp->val) {
		lists[i] = lists[smallChild];
		i = smallChild;
		smallChild = 2 * i + 1;
		if (smallChild + 1 < size && lists[smallChild]->val > lists[smallChild + 1]->val)
			++smallChild;
	}
	lists[i] = tmp;
}
//堆實現優先隊列，k個鏈表首元素建堆，每次彈出堆頂元素；
//刪除堆頂元素後重構需要log(k)個操作，總時間複雜度n*log(k)
struct ListNode* mergeKLists(struct ListNode** lists, int listsSize) {
	if (lists == NULL || listsSize <= 0)
		return NULL;
	if (listsSize == 1)
		return lists[0];
	int size;
	struct ListNode* head = (struct ListNode *)malloc(sizeof(struct ListNode));	//啞節點
	struct ListNode* p = head;
	head->next = NULL;					//不可少，否則所有鏈表爲空時無法返回NULL
	size = buildHeap(lists, listsSize);	//將lists按堆序排序,並返回初始堆大小，即非空鏈表數量
	while (size) {						//只要堆非空，每次彈出堆頂元素，更新堆
		p->next = lists[0];
		p = p->next;
		lists[0] = lists[0]->next;		//更新堆頂元素
		if (lists[0] == NULL) {			//如果堆頂元素爲空，將尾部元素移至堆頂
			lists[0] = lists[size - 1];
			--size;
		}
		percolateDown(lists, size);		//下濾堆頂元素使其保持堆序性
	}
	p = head->next;
	free(head);
	return p;
}