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) {}
 * };
 */
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
     */
    // 解题思路，将k个链表拆分，两两合并
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        // write code here
        if (lists.size() == 0) {
            return nullptr;
        }
        if (lists.size() == 1) {
            return lists[0];
        }
        ListNode* head = lists[0];
        for(int i = 1; i < lists.size(); ++i) {
            head = MergeList(head, lists[i]);
        }
        return head;
    }

    // 升序合并两个链表
    ListNode* MergeList(ListNode* pHead1, ListNode* pHead2)
    {
        if (!pHead1) {
            return pHead2;
        }
        if (!pHead2) {
            return pHead1;
        }
        ListNode* head = new ListNode(0);
        ListNode* cur = head;
        while (pHead1 && pHead2) {
            if (pHead1->val <= pHead2->val) {
                cur->next = pHead1;
                pHead1 = pHead1->next;
            } else {
                cur->next = pHead2;
                pHead2 = pHead2->next;
            }
            cur = cur->next;
        }
        if (pHead1) {
            cur->next = pHead1;
        }
        if (pHead2) {
            cur->next = pHead2;
        }
        return head->next;
    }
};