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(vector<ListNode*>& lists) {
        // write code here
        ListNode* head = nullptr;
        int k = lists.size();
        if (k >= 2) {
            int mid = k / 2;
            std::vector<ListNode*> leftLists;
            leftLists.resize(mid);
            std::vector<ListNode*> rightLists;
            rightLists.resize(k-mid);
            copy(lists.begin(), lists.begin() + mid, leftLists.begin());
            copy(lists.begin() + mid, lists.end(), rightLists.begin());
            return merge2Lists(mergeKLists(leftLists), mergeKLists(rightLists));
        }else if (k == 1) {
            return lists[0];
        }
        return nullptr;
    }

    ListNode* merge2Lists(ListNode* left, ListNode* right) {
        if (left == nullptr) {
            return right;
        }
        if (right == nullptr) {
            return left;
        }
        ListNode* head = nullptr;
        ListNode* currentHead = nullptr;
        while (left != nullptr and right != nullptr) {
            ListNode* smallerNode = left->val <= right->val ? left : right;
            if (currentHead != nullptr) {
                currentHead->next = smallerNode;
            } else {
                head = smallerNode;
            }
            currentHead = smallerNode;
            if (left->val <= right->val) {
                left = left->next;
            } else {
                right = right->next;
            }
        }
        if (left == nullptr) {
            currentHead->next = right;
        } else {
            currentHead->next = left;
        }
        return head;
    }
};