LeetCode（23）Merge k Sorted Lists

题目

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

分析

合并k个有序链表。

我们从数据结构的链表章节学会了如何合并两个链表，针对此题，一个简单的解法便是遍历k次，两个链表合并，合并后的结果与下一个链表合并；

此时时间复杂度为O(nk)，遗憾，提交会出现TLE。

所以，必须另觅他法，我们知道排序领域中很多方法都可以提高性能比如合并排序性能为O(nlogn)；可以应用到此题，可以顺利AC，时间复杂度为O(nlogk)

AC代码

class Solution {
public:
    /*方法一：采用合并排序的思路 复杂度为O(nlogk)*/
    ListNode* mergeKLists(vector<ListNode*>& lists) {
        int len = lists.size();
        ListNode *head = NULL;
        if (len == 0)
            return head;
        else if (len == 1)
        {
            head = lists[0];
            return head;
        }
        else{
            int mid = (len - 1) / 2;
            vector<ListNode *> l1(lists.begin(), lists.begin() + mid + 1);
            vector<ListNode *> l2(lists.begin() + mid + 1, lists.end());
            ListNode *head1 = mergeKLists(l1);
            ListNode *head2 = mergeKLists(l2);

            return mergeTwoLists(head1, head2);
        }//else
    }

    /*方法二：暴力法 复杂度为O(nK) 出现超时*/
    ListNode* mergeKLists2(vector<ListNode*>& lists) {
        int len = lists.size();
        ListNode *head = NULL;
        if (len == 0)
            return head;
        else if (len == 1)
        {
            head = lists[0];
            return head;
        }
        else{
            head = mergeTwoLists(lists[0], lists[1]);
            for (int i = 2; i < len; i++)
                head = mergeTwoLists(head, lists[i]);
        }
        return head;
    }

    /*合并两个链表函数*/
    ListNode* mergeTwoLists(ListNode* l1, ListNode* l2) {
        if (l1 == NULL)
            return l2;
        if (l2 == NULL)
            return l1;

        //合并后链表初始化为空
        ListNode *rl = NULL;

        ListNode *p = l1, *q = l2;
        if (l1->val <= l2->val)
        {
            rl = l1;
            p = l1->next;
        }
        else{
            rl = l2;
            q = l2->next;
        }
        rl->next = NULL;
        ListNode *head = rl;

        while (p && q)
        {
            if (p->val <= q->val)
            {
                rl->next = p;
                p = p->next;
            }
            else{
                rl->next = q;
                q = q->next;
            }//else
            rl = rl->next;
        }//while

        while (p)
        {
            rl->next = p;
            p = p->next;
            rl = rl->next;
        }

        while (q)
        {
            rl->next = q;
            q = q->next;
            rl = rl->next;
        }

        return head;
    }
};

GitHub测试程序源码