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 <cstddef>
class Solution {
public:

    ListNode* Merge(ListNode* pHead1, ListNode* pHead2)
    {
        ListNode* pDummy = new ListNode(-1);
        ListNode* pCurr = pDummy;
        
        while (pHead1 && pHead2) {
            if (pHead1->val < pHead2->val) {
                pCurr->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                pCurr->next = pHead2;
                pHead2 = pHead2->next;
            }
            
            pCurr = pCurr->next;
        }
        pCurr->next = pHead1 ? pHead1 : pHead2;

        return pDummy->next;
    }

    ListNode* devideMerge(vector<ListNode*>& lists, int left, int right) {

        if (left > right) {
            return NULL;
        }
	   
	    if (left == right) {
            return lists[left];
        }

        int mid = (left + right) / 2;
        return Merge(devideMerge(lists, left, mid), devideMerge(lists, mid + 1, right));
    }

    /**
     * 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 devideMerge(lists, 0, lists.size() - 1);
    }
};