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 <vector>
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* Merge(ListNode* pHead1, ListNode* pHead2) {
        // write code here
        if (pHead1 == nullptr) {
            return pHead2;
        }
        if (pHead2 == nullptr) {
            return pHead1;
        }
        ListNode* head = pHead1;
        int temp1 = pHead1->val;
        int temp2 = pHead2->val;
        if (temp1 > temp2) {
            head = pHead2;
            pHead2 = pHead2->next;
        } else {
            head = pHead1;
            pHead1 = pHead1->next;
        }
        ListNode* tempNode = head;
        while (pHead1 != nullptr && pHead2 != nullptr) {
            temp1 = pHead1->val;
            temp2 = pHead2->val;
            if (temp1 > temp2) {
                tempNode->next = pHead2;
                tempNode = pHead2;
                pHead2 = pHead2->next;
            } else {
                tempNode->next = pHead1;
                tempNode = pHead1;
                pHead1 = pHead1->next;
            }
        }
        if (pHead1 == nullptr) {
            tempNode->next = pHead2;
        } else {
            tempNode->next = pHead1;
        }

        return head;
    }
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        ListNode* head = nullptr;
        if (!lists.empty()) {
            int N=lists.size();
            while (N>1) {
                for (int i=0; i<N/2; i++) {
                    lists[i]= Merge(lists[i],lists[N-i-1]);
                }
                if (N%2==1) {
                    N=N/2+1;
                }else {
                    N=N/2;
                }
            }
            head = lists[0];
        }
        return head;
    }
};

写这题之前先去完成这道题：https://www.nowcoder.com/share/jump/4047987181733831176674

合并两个链表会了之后，两两合并这些链表即可。由于两两合并，时间复杂度为log(n)*n