Last Updated on 2021-07-20 by Clay
題目
You are given an array of k
linked-lists lists
, each linked-list is sorted in ascending order.
Merge all the linked-lists into one sorted linked-list and return it.
Example 1:
Input: lists = [[1,4,5],[1,3,4],[2,6]]
Output: [1,1,2,3,4,4,5,6]
Explanation: The linked-lists are:
[
1->4->5,
1->3->4,
2->6
]
merging them into one sorted list:
1->1->2->3->4->4->5->6
Example 2:
Input: lists = []
Output: []
Example 3:
Input: lists = [[]]
Output: []
Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.
Constraints:
k == lists.length
0 <= k <= 10^4
0 <= lists[i].length <= 500
-10^4 <= lists[i][j] <= 10^4
lists[i]
is sorted in ascending order.- The sum of
lists[i].length
won't exceed10^4
.
解題思路
這個題目我嘗試的解題方法比較暴力,或許沒什麼參考性也說不一定。
(處理愛貓生病,暫時停止撰寫內文,純做紀錄)
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode() : val(0), next(nullptr) {} * ListNode(int x) : val(x), next(nullptr) {} * ListNode(int x, ListNode *next) : val(x), next(next) {} * }; */ class Solution { public: ListNode* mergeKLists(vector<ListNode*>& lists) { vector<int> v; // Read all list for (int i=0; i<lists.size(); i++) { ListNode* readNode = new ListNode(0, lists[i]); while (readNode->next != NULL) { v.push_back(readNode->next->val); readNode->next = readNode->next->next; } } // Sort sort(v.begin(), v.end()); // Create ListNode* head = new ListNode(); head->val = -1; ListNode* currentNode = new ListNode(0, head); for (int i=0; i<v.size(); i++) { ListNode* newNode = new ListNode(); currentNode->next->val = v[i]; if (i != v.size()-1) { ListNode* head = new ListNode(); currentNode->next->next = newNode; currentNode->next = currentNode->next->next; } else { currentNode->next->next = NULL; } } cout << head->val << endl; // Excepted if (head->val == -1 and head->next == NULL) { return NULL; } return head; } };