Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity 1. Naive Solution 思路:直接的想法依次找出列表中的最小项串接起来。 复杂度分析:时间复杂度:O(k*n), 空间复杂度:O(c)。其中 n为最长子列表长度,c为常量。 ListNode *mergeKLists(vector<ListNode *>&l...
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 合并k个有序链表。 二、解题报告 解法一:两两合并 由于前面刚刚做过《LeetCode 21 - Merge Two Sorted Lists》,看到这个题的第一反应就是两两合并,还可以直接调用mergeTwoLists()。 classSolution{public:...
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 题意:将k条已排序的链表合并为一条排序的链表; 解法:将多条链表合并为一条有序的链表,其实与两条...
class Solution { public: ListNode* mergeKLists(vector<ListNode*>& lists) { if(lists.size()==0){return NULL;} if(lists.size()==1){return lists[0];} queue<ListNode*> que(deque(lists.begin(),lists.end())); while(que.size()>1){ ListNode* p1 =que.front(); que.pop(); ListNode...
之前做过两个有序链表的排序插入Leetcode21 Merge Two Sorted Lists。当时有暴力循环迭代的方法,还有递归的方法。每次加入一个元素,然后对剩下的元素继续调用函数进行排序插入。 这次是k个,感觉总体思路不会差太多。如果我们想要继续使用递归的方法,对于参数的处理就不会合并两个链表一样简单。因为当时只有两个参数,...
2. Solution Version 1 /** * 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){intsize=lists.size();if(size==0){returnNULL;...
class Solution { public: void heapify(vector<int> &arr, int i, int n) { int largest = i; int left = 2 * i + 1; int right = 2 * i + 2; if (left < n && arr[left] > arr[largest]) { largest = left; } if (right < n && arr[right] > arr[largest]) { largest = ...
Can you solve this real interview question? Merge k Sorted Lists - 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: Inpu
LeetCode 23. Merge k Sorted Lists 经典题 解法一:优先队列 {{{ class CMP{ public: bool operator () (const ListNode* a, const ListNode * b) const { return a->val > b->val; } }; class Solution { public: ListNode* mergeKLists(vector<ListNode*>& lists) { ListNode *head = nullptr;...
Mergeksorted linked lists and return it as one sorted list. Analyze and describe its complexity. 合并k个有序的链表,我们假设每个链表的平均长度是n。这一题需要用到合并两个有序的链表子过程 算法1: 最傻的做法就是先1、2合并,12结果和3合并,123结果和4合并,…,123..k-1结果和k合并,我们计算一下...