Description
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
多路排序的一个例子,用优先权队列保存指针,每次选出有最小的权值的指针,将其指向的节点加入结果链表。
努力改良自己代码风格,让自己的代码变得好看。
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: struct cmp { bool operator()(ListNode *a, ListNode *b) { return a->val > b->val; } }; priority_queue<ListNode*, vector<ListNode* >, cmp > pq; ListNode* mergeKLists(vector<ListNode*>& lists) { for(int i = 0; i < lists.size(); i++) { if(lists[i] != NULL) { pq.push(lists[i]); } } ListNode *head = NULL; ListNode *p = NULL; while(!pq.empty()) { ListNode *tmp = pq.top(); pq.pop(); if(head == NULL) { head = new ListNode(tmp->val); p = head; } else { ListNode *t = new ListNode(tmp->val); p->next = t; p = t; } tmp = tmp->next; if(tmp != NULL) pq.push(tmp); } return head; } };