problem:
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. Tags Divide and Conquer Linked List Heap
合并K个已序单链表
thinking:
(1)题目没有要求不能够新开ListNode,所以暴力破解法:提取K个list的keyword。排序、新建结点插入。这样的情况对原list是否排好序没有要求。
排序时间复杂度能够做到O(N* log N )。提取keyword和新建结点的时间复杂度都为O(N),所以总的时间复杂度为O(N*logN),这没有考虑新建结点的时间浪费和空间 浪费。
(2)採用能够容纳结点的容器,想到的是堆,或者堆的封装-优先队列,因为堆的O(N*logN)排序的时间复杂度。并且不用新开结点,节省空间。
暴力法:
/** * 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) { ListNode *newlist=NULL; vector<int> tem; for(int i=0;i<lists.size();i++) { while(lists.at(i)!=NULL) { tem.push_back(lists.at(i)->val); lists.at(i)=lists.at(i)->next; } } if(tem.size()==0) return NULL; sort(tem.begin(),tem.end()); for(int i=tem.size()-1;i>=0;i--) { ListNode *p = new ListNode(tem.at(i)); p->next = newlist; newlist = p; } return newlist; }//mergeKLists() };
优先队列法:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class ListNodeCompare:public binary_function<ListNode*,ListNode*,bool> { public: bool operator()(ListNode* t1,ListNode* t2)const { if ( !t1||!t2 ) return !t2; return t1->val>t2->val; } }; class Solution { public: ListNode *mergeKLists(vector<ListNode *> &lists) { // Note: The Solution object is instantiated only once and is reused by each test case. if (lists.empty()) return NULL; priority_queue<ListNode*,vector<ListNode*>,ListNodeCompare> Q; for(int i=0;i<lists.size();i++) if ( lists[i]!=NULL) Q.push(lists[i]); ListNode guard(-1); ListNode* tail=&guard; while(!Q.empty()) { ListNode* toAdd=Q.top(); Q.pop(); tail->next=toAdd; tail=tail->next; if (toAdd->next) Q.push(toAdd->next); } return guard.next; } };
时间: 2024-11-05 15:53:28