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) {}
 * };
 */
#include <list>
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(vector<ListNode*>& lists) {
        // write code here
        if (lists.empty()) {
            return nullptr;
        }
        if (lists.size() < 2) {
            return lists[0];
        }
        std::list<ListNode*> l;
        for (auto p : lists) {
            l.push_back(p);
        }
        while (l.size() > 1) {
            ListNode* p1 = l.front();
            l.pop_front();
            ListNode* p2 = l.front();
            l.pop_front();
            l.push_back(Merge2(p1, p2));
        }

        return l.front();
    }
private:
    ListNode* Merge2(ListNode* p1, ListNode* p2) {
        if (!p1) {
            return p2;
        }
        if (!p2) {
            return p1;
        }

        auto* pre = new ListNode(-1);
        auto* new_head = pre;
        pre->next = p1;
        while (p1 && p2) {
            if (p1->val < p2->val) {
                pre->next = p1;
                p1 = p1->next;
            } else {
                pre->next = p2;
                p2 = p2->next;
            }
            pre = pre->next;
        }
        if (p1) {
            pre->next = p1;
        }
        if (p2) {
            pre->next = p2;
        }
        return new_head->next;
    }
};