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 <iostream>
class Solution {
  public:
    ListNode* mergeTwo(ListNode* nodes1, ListNode* nodes2) {
        ListNode* newNodes;
        if (nodes1 == nullptr) {
            newNodes = nodes2;
            return newNodes;
        } else if (nodes2 == nullptr) {
            newNodes = nodes1;
            return newNodes;
        }

        if (nodes1->val <= nodes2->val) {
            newNodes = nodes1;
            nodes1 = nodes1->next;
        } else {
            newNodes = nodes2;
            nodes2 = nodes2->next;
        }
        ListNode* headNodes = newNodes;
        while (true) {
            if (nodes1 == nullptr) {
                newNodes->next = nodes2;
                break;
            } else if (nodes2 == nullptr) {
                newNodes->next = nodes1;
                break;
            } else if (nodes1->val <= nodes2->val) {
                newNodes->next = nodes1;
                newNodes = newNodes->next;
                nodes1 = nodes1->next;
            } else {
                newNodes->next = nodes2;
                newNodes = newNodes->next;
                nodes2 = nodes2->next;
            }
        }

        return headNodes;
    }
    /**
     * 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* newNodes = nullptr;
        for (auto& list : lists) {
            newNodes = mergeTwo(newNodes, list);
        }

        return newNodes;
    }
};