LeetCode 023. Merge k Sorted Lists

Merge k Sorted Lists

 

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

一开始的思路是先将两个链表合并，然后每次从剩下的链表中取出一个链表与之合并。假设共有m条链表且每条链表长度均为n，则复杂度为：

2n+3N+4n+5n+6n+……+mn = (m-1)(m+2)n/2

也即复杂度为O(n*m^2)

经过验证发现超时了。因此，修改了思路。借鉴堆排序思想，构建一个小顶堆，每次从堆顶中取一个元素构建链表，并加入一个新的元素重新建堆。该思路算法复杂度为O(mn)

代码如下：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public://已知heap_array中除了heap_array[s]外均满足小顶堆的定义//调整heap_array[s]使heap_array成为一个小顶堆void HeapAdjust(vector<ListNode *> & heap_array, int s, int n){ListNode * temp = heap_array[s];for(int i=2*s; i<=n; i*=2){if(i<n && heap_array[i]->val > heap_array[i+1]->val)i++;if(temp->val <= heap_array[i]->val)break;heap_array[s] = heap_array[i];s = i;}heap_array[s] = temp;}    ListNode *mergeKLists(vector<ListNode *> &lists)     {int num = lists.size();int length = 0; //lists中非空链表数目int i = 0;if(!num)return NULL;//将非空链表置于lists前面for(i=0; i<num; i++){if(lists[i]!=NULL){if(i != length)lists[length] = lists[i];length ++;}}if(!length)return NULL;else if(length == 1)return lists[0];ListNode * head = new ListNode(0);ListNode * travel = head;//构建一个数组，建堆并进行堆排序,堆排序，有效数据下标从1开始vector <ListNode *> heap_array (length+1);for(i=1; i<=length; i++)heap_array[i] = lists[i-1];//建堆for(i=length/2; i>0; i--)HeapAdjust(heap_array,i,length);//取出最小结点，并更新数组while(length > 1){travel->next = heap_array[1];travel = travel->next;if(heap_array[1]->next != NULL)heap_array[1] = heap_array[1]->next;else  //如果某一个链表已经完全取尽，则将最后一个链表换上，并且length-1{heap_array[1] = heap_array[length];heap_array[length] = NULL;length = length -1;}HeapAdjust(heap_array,1,length);}travel->next = heap_array[1];travel = head;head = head->next;delete travel;return head;    }};