标签:div sts head nbsp vector nullptr res new lin
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
Example: Input: [ 1->4->5, 1->3->4, 2->6 ] Output: 1->1->2->3->4->4->5->6
合并k个有序链表。
用归并排序就好了。
1 class Solution { 2 public: 3 ListNode* merge(vector<ListNode*>& lists, int low, int high) { 4 ListNode* left = nullptr; 5 ListNode* right = nullptr; 6 if (low < high - 1) { 7 int mid = low + (high - low) / 2; 8 left = merge(lists, low, mid); 9 right = merge(lists, mid+1, high); 10 } 11 else { 12 left = lists[low]; 13 right = low == high? nullptr: lists[high]; 14 } 15 int a, b; 16 ListNode *res = new ListNode(0); 17 ListNode *curr = res; 18 while (left || right) { 19 a = left? left->val: INT_MAX; 20 b = right? right->val: INT_MAX; 21 if (a < b) { 22 curr->next = left; 23 left = left->next; 24 curr = curr->next; 25 } 26 else { 27 curr->next = right; 28 right = right->next; 29 curr = curr->next; 30 } 31 } 32 curr = res->next; 33 delete res; 34 return curr; 35 } 36 ListNode* mergeKLists(vector<ListNode*>& lists) { 37 ListNode *head = nullptr; 38 int low = 0; 39 int high = lists.size() - 1; 40 if (low <= high) 41 head = merge(lists, low, high); 42 return head; 43 } 44 };
标签:div sts head nbsp vector nullptr res new lin
原文地址:https://www.cnblogs.com/Zzz-y/p/9063585.html