The basic idea is to first scan the list, find the middle point and break the list into two, sort two sub-lists recursively and merge them together. Obviously, time complexity would be O(nlogn). What is the spac
Merge k sorted linked list就是merge 2 sorted linked list的变形题。 而且我们很自然的就想到了经典的Merge Sort,只不过那个是对数组进行sort。而不同的地方,仅仅是Merge两个list的操作不同。 这里来复习一下Merge Sort(对于数组操作),参考Wikipedia: 归并操作(merge),也叫归并算法,指的是将两个已经排序的序列...
intN, n, temp; element_t* node = list->head; N = list_length( list );// Don't sort an unsortable listif(N < 2)return;// just swap the two elements if necessaryif(N == 2) {if(list->head->val > list->tail->val) { temp = list->head->val; list->head->val = list...
Merge all the linked-lists into one sorted linked-list and return it. Example 1: Input:lists = [[1,4,5],[1,3,4],[2,6]]Output:[1,1,2,3,4,4,5,6]Explanation:The linked-lists are: [ 1->4->5, 1->3->4, 2->6 ] merging them into one sorted list: 1->1->2->3->4...
Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. 题解: 是Merge Two Sorted Lists的进阶版,现在是k条,用divide and conquer, 两条两条合并 最后合并成一个大的。 Note: mergeSort中别忘了最后的返回值是lists[l]. ...
I am facing a runtime issue in my code.Can anyone help me with my code: My code link:https://pastebin.com/qCC4GsPS. Just check the merge and mergesort function in this link. #merge sort,#linked list -6 rsudhanshu138 4 years ago ...
vector<int>v; for(inti=0;i<lists.size();i++) for(ListNode*p=lists[i];p!=NULL;p=p->next)v.push_back(p->val); if(v.size()==0)returnnullptr; sort(v.begin(),v.end()); vector<int>::iteratorit; ListNode*p1=(ListNode*)malloc(sizeof(ListNode)); ...
这样下一次只需合并3个链表,我们再合并1和3,最后和2合并就可以了。 总结 关于堆,理解的不是很透彻,可以参考以下两个文章继续学习:http://bubkoo.com/2014/01/14/sort-algorithm/heap-sort/https://github.com/qiwsir/algorithm/blob/master/heapq.md...
Merge sort functions by partitioning the input into smaller sub-arrays, sorting each sub-array recursively, and subsequently merging the sorted sub-arrays.
sort(nodeVal.begin(),nodeVal.end()); boolisFirst=1; ListNode*res=nullptr,*head=nullptr; for(autop:nodeVal) { if(isFirst) { isFirst=0; head=newListNode(p); res=head; } else { head->next=newListNode(p); head=head->next;