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
        return divMerge(lists, 0, lists.size()-1);
    }

    ListNode * divMerge(vector<ListNode*>& lists, int start, int end) {

        if (start > end) {
            return nullptr;
        }else if(start == end) {
            return lists[start];
        }

        int mid = start + (end - start) / 2;

        return merge(divMerge(lists, start, mid), divMerge(lists, mid + 1, end));
    }

    ListNode * merge(ListNode* pHead1, ListNode *pHead2) {

        if (pHead1 == nullptr) {
            return pHead2;
        }

        if (pHead2 == nullptr) {
            return pHead1;
        }

        ListNode * vhead = new ListNode(-1);

        ListNode * curr = vhead;

        while(pHead1 && pHead2) {
            if (pHead1->val <= pHead2->val) {
                curr->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                curr->next = pHead2;
                pHead2 = pHead2->next;
            }
            curr = curr->next;
        }
        curr->next = pHead1 ? pHead1 : pHead2;

        ListNode* tmp = vhead;
        vhead = vhead->next;
        delete tmp;
        return vhead;
    }
};