合并k个已排序的链表_牛客题霸_牛客网 (nowcoder.com)

class Solution {
  public:
    //两个有序链表合并函数
    ListNode* Merge2(ListNode* pHead1, ListNode* pHead2) {
        if(!pHead1) return pHead2;
        if(!pHead2) return pHead1;
        if(pHead1->val < pHead2->val){
            pHead1->next = Merge2(pHead1->next,pHead2);
            return pHead1;
        }else{
            pHead2->next = Merge2(pHead2->next,pHead1);
            return pHead2;
        }
    }
    //划分合并区间函数
    /*
    step 1：从链表数组的首和尾开始，每次划分从中间开始划分，划分成两半，得到左边
    n/2个链表和右边 n/2个链表。
    step 2：继续不断递归划分，直到每部分链表数为1.
    */
    ListNode* divideMerge(vector<ListNode*>& lists, int left, int right) {
         if(left > right) return nullptr;
         if(left == right) return lists[left];
         int mid = left + (right - left)/2;
         return Merge2(divideMerge(lists,left,mid),divideMerge(lists,mid+1,right));
    }
    
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        //k个链表归并排序
        return divideMerge(lists, 0, lists.size() - 1);
    }
};