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