Leetcode:23. Merge k Sorted Lists (week 10)

Description

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

解题思路

利用分治的思想把合并k个链表分成两个合并k/2个链表的任务，一直划分，知道任务中只剩一个链表或者两个链表。可以很简单的用递归来实现。因此算法复杂度为T(k) = 2T(k/2) + O(nk),算法复杂度为O（nklogk）

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* mergeKLists(vector<ListNode*>& lists) {        int len = lists.size();        if (len == 0) return NULL;        if (len == 1) return lists[0];        return merge(lists,0, lists.size()-1);    }    ListNode* merge(vector<ListNode*>& lists,int start,int end) {        if (start == end) return lists[start];        else if (start+1 == end) return merge2(lists[start], lists[end]);        ListNode* l1 = merge(lists,start, (start+end)/2);        ListNode* l2 = merge(lists,(start+end)/2+1,end);        return merge2(l1,l2);    }    ListNode* merge2(ListNode* l1, ListNode* l2) {        if(l1==l2) return l1;        if (!l1) return l2;        if (!l2) return l1;        if (l1->val > l2->val) return merge2(l2,l1);        ListNode* newl2= new ListNode(0);        newl2->next = l2;        ListNode * p1 = l1;        while (p1->next && newl2->next) {            if (p1->next->val < newl2->next->val) {                p1 = p1->next;            } else{                ListNode * temp = p1->next;                p1->next = newl2->next;                newl2->next = newl2->next->next;                p1->next->next = temp;                p1= p1->next;            }        }        if (!p1->next) p1->next = newl2->next;        delete newl2;        return l1;    }};