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* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        return divideMerge(lists, 0, lists.size() - 1);
    }
    ListNode* divideMerge(vector<ListNode*>& lists, int left, int right) {
        if (left > right) {
            return nullptr;
        } else if (left == right) {
            return lists[left];
        } else {
            int mid = (left + right) / 2;
            return Merge2(divideMerge(lists, left, mid), divideMerge(lists, mid + 1,
                          right));
        }

    }
    ListNode* Merge2(ListNode* pHead1, ListNode* pHead2) {
        //一个已经为空了，直接返回另一个
        if (pHead1 == NULL)
            return pHead2;
        if (pHead2 == NULL)
            return pHead1;
        //加一个表头
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        //两个链表都要不为空
        while (pHead1 && pHead2) {
            //取较小值的节点
            if (pHead1->val <= pHead2->val) {
                cur->next = pHead1;
                //只移动取值的指针
                pHead1 = pHead1->next;
            } else {
                cur->next = pHead2;
                //只移动取值的指针
                pHead2 = pHead2->next;
            }
            //指针后移
            cur = cur->next;
        }
        //哪个链表还有剩，直接连在后面
        if (pHead1)
            cur->next = pHead1;
        else
            cur->next = pHead2;
        //返回值去掉表头
        return head->next;
    }
};