Merge k Sorted Lists
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
以前做的ac,结果前几天提交发现超时了,原来系统最近又加了一个test case。。。用的选择排序,时间复杂度是O(n^2)
采用最小堆,首先将 k 个首节点放入堆中,弹出最小的节点并插入到新的链表中;
弹出的节点如果next 非空,就将它的 下一个节点进 堆。
继续,直到堆为空。
堆 Push 和 pop 的复杂度是 log(k), 所以复杂度是 nlg(k), n为总的节点数
/** * 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 K = lists.size(); if (K == 0) return NULL; else if (K == 1) return lists[0]; ListNode *listHead(NULL), *listRear(NULL); ListNode *node(NULL); priority_queue<ListNode*, vector<ListNode*>, cmp> h; // push K list heads into heap for(int i=0; i<K; ++i) if(lists[i] != NULL){ h.push(lists[i]); lists[i] = lists[i]->next; } while(!h.empty()){ //pop the min of k nodes node = h.top(); h.pop(); if(node->next) h.push(node->next); //insert node into new list if(listRear){ listRear->next = node; listRear = listRear->next; } else{ listHead = listRear = node; } } return listHead; } private: struct cmp{ bool operator()(ListNode* lhs, ListNode *rhs){ if(lhs->val < rhs->val) return false; else return true; } }; };
http://blog.csdn.net/shoulinjun/article/details/19576585