题解 | #合并k个已排序的链表#

/**
 * struct ListNode {
 *	int val;
 *	struct ListNode *next;
 *	ListNode(int x) : val(x), next(nullptr) {}
 * };
 */
class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param lists ListNode类vector 
     * @return ListNode类
     */
    ListNode* mergeKLists(std::vector<ListNode*>& lists) {

			if (!lists.size()) return nullptr;
			ListNode LL = ListNode(1111);
			ListNode* PreHead = &LL, * *temp = nullptr, * pHead = &LL;
			while (true) {
				int count = 1111;
				for (int i = 0; i < lists.size(); i++) {

					if (lists[i]) {
						if (count > lists[i]->val) {
							count = lists[i]->val;
							// Find least Node
							temp = &lists[i];
							std::cout << "count =" << count << std::endl;
						}
					}
					
				}
				if (count == 1111) return PreHead->next;
				if (temp) *temp = (*temp)->next;
				pHead->next = new ListNode(count);
				pHead = pHead->next;

			}


			return PreHead->next;


	}
};