Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
合並k個有序的鏈表,我們假設每個鏈表的平均長度是n。這一題需要用到合並兩個有序的鏈表子過程
算法1:
最傻的做法就是先1、2合並,12結果和3合並,123結果和4合並,…,123..k-1結果和k合並,我們計算一下復雜度。
1、2合並,遍歷2n個節點
12結果和3合並,遍歷3n個節點
123結果和4合並,遍歷4n個節點
…
123..k-1結果和k合並,遍歷kn個節點
總共遍歷的節點數目為n(2+3+…+k) = n*(k^2+k-2)/2, 因此時間復雜度是O(n*(k^2+k-2)/2) = O(nk^2),代碼如下:
/** * 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) { if(lists.size() == 0)return NULL; ListNode*res = lists[0]; for(int i = 1; i < lists.size(); i++) res = merge2list(res, lists[i]); return res; } ListNode *merge2list(ListNode *head1, ListNode*head2) { ListNode node(0), *res = &node; while(head1 && head2) { if(head1->val <= head2->val) { res->next = head1; head1 = head1->next; } else { res->next = head2; head2 = head2->next; } res = res->next; } if(head1) res->next = head1; else if(head2) res->next = head2; return node.next; } };
算法2:利用分治的思想把合並k個鏈表分成兩個合並k/2個鏈表的任務,一直划分,知道任務中只剩一個鏈表或者兩個鏈表。可以很簡單的用遞歸來實現。因此算法復雜度為T(k) = 2T(k/2) + O(nk),很簡單可以推導得到算法復雜度為O(nklogk)
遞歸的代碼就不貼了。下面是非遞歸的代碼非遞歸的思想是(以四個鏈表為例): 本文地址
1、3合並,合並結果放到1的位置
2、4合並,合並結果放到2的位置
再把1、2合並(相當於原來的13 和 24合並)
/** * 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 n = lists.size(); if(n == 0)return NULL; while(n >1) { int k = (n+1)/2; for(int i = 0; i < n/2; i++) lists[i] = merge2list(lists[i], lists[i + k]); n = k; } return lists[0]; } ListNode *merge2list(ListNode *head1, ListNode*head2) { ListNode node(0), *res = &node; while(head1 && head2) { if(head1->val <= head2->val) { res->next = head1; head1 = head1->next; } else { res->next = head2; head2 = head2->next; } res = res->next; } if(head1) res->next = head1; else if(head2) res->next = head2; return node.next; } };
算法3:維護一個k個大小的最小堆,初始化堆中元素為每個鏈表的頭結點,每次從堆中選擇最小的元素加入到結果鏈表,再選擇該最小元素所在鏈表的下一個節點加入到堆中。這樣當堆為空時,所有鏈表的節點都已經加入了結果鏈表。元素加入堆中的復雜度為O(longk),總共有kn個元素加入堆中,因此,復雜度也和算法2一樣是O(nklogk)
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { private: struct cmp { bool operator ()(const ListNode *a, const ListNode *b) { return a->val > b->val; } }; public: ListNode *mergeKLists(vector<ListNode *> &lists) { int n = lists.size(); if(n == 0)return NULL; ListNode node(0), *res = &node; priority_queue<ListNode*, vector<ListNode*>, cmp> que; for(int i = 0; i < n; i++) if(lists[i]) que.push(lists[i]); while(!que.empty()) { ListNode * p = que.top(); que.pop(); res->next = p; res = p; if(p->next) que.push(p->next); } return node.next; } };
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