[Leetcode] #23 Merge k Sorted Lists

Discription:

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

Solution:

ListNode* merge(ListNode *left, ListNode *right){ListNode node(-1);ListNode *cur = &node;while (left && right){if (left->val < right->val){cur->next = left;left = left->next;}else{cur->next = right;right = right->next;}cur = cur->next;}cur->next = left ? left : right;return node.next;}ListNode* mergeKLists(vector<ListNode*>& lists, int begin, int end) {if (begin == end)return lists[begin];int mid = (begin + end) >> 1;ListNode *left = mergeKLists(lists, begin, mid);ListNode *right = mergeKLists(lists, mid + 1, end);return merge(left, right);}ListNode* mergeKLists(vector<ListNode*>& lists) {if (lists.empty())  return NULL;ListNode *head = mergeKLists(lists, 0, lists.size() - 1);return head;}

//方法二ListNode *mergeKLists(vector<ListNode *> &lists) {if (lists.empty()){return nullptr;}while (lists.size() > 1){lists.push_back(mergeTwoLists(lists[0], lists[1]));lists.erase(lists.begin());lists.erase(lists.begin());}return lists.front();}ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {if (l1 == nullptr){return l2;}if (l2 == nullptr){return l1;}if (l1->val <= l2->val){l1->next = mergeTwoLists(l1->next, l2);return l1;}else{l2->next = mergeTwoLists(l1, l2->next);return l2;}}