有序链表的合并

/*
LeetCode_0023
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
*/

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

1.以递归的方式

ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
		if (l1 == NULL)
			return l2;
		if (l2 == NULL)
			return l1;
		if (l1->val <= l2->val) {
			l1->next = mergeTwo(l1->next, l2);
			return l1;
		}
		if (l1->val > l2->val) {
			l2->next = mergeTwo(l1, l2->next);
			return l2;
		}
	}

2.直接合并

	ListNode* mergeTwo(ListNode* l1, ListNode* l2) {
		ListNode dummy(-1);
		ListNode* result = &dummy;
		while (l1&&l2) {
			if (l1->val <= l2->val) {
				result->next = l1;
				result = result->next;
				l1 = l1->next;
			}
			else if (l1->val > l2->val) {
				result->next = l2;
				result = result->next;
				l2 = l2->next;
			}
		}
		while (l1)
			result->next = l1;
		while (l2)
			result->next = l2;
		return dummy.next;
	}

3.合并k个有序链表

ListNode *mergeKLists(vector<ListNode *> &lists) {
	if (lists.empty()) {
		return nullptr;
	}
	while (lists.size() > 1) {
		lists.push_back(mergeTwoLists(lists[0], lists[1]));
		lists.erase(lists.begin());
		lists.erase(lists.begin());
	}
	return lists.front();
}

ListNode *mergeKLists(vector<ListNode *> &lists) {
	if (lists.empty()) return NULL;
        int len = lists.size();
        while (len > 1) 
        {
            for (int i = 0; i < len / 2; ++i) 
            {
                lists[i] = mergeTwoLists(lists[i], lists[len - 1 - i]);
            }
            len = (len + 1) / 2;
        }
        return lists.front();
}