这是小川的第413次更新,第446篇原创 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第264题(顺位题号是1175).返回1到n的排列数,以使质数处于质数索引(索引从1开始).(请记住,当且仅当整数大于1,并且不能将其写为两个均小于它的正整数的乘积,它才是质数.)由于答案可能很大,因此请以10^9 + 7为模返回答案. 例如: 输入:n = 5 输出:12 说明:[1,2,5,4,3]是有效的排列,但是[5,2,3,4,1]并不是,因为素数5在索引1处. 输入:n = 100

LeetCode 25 k组一个翻转链表 TITLE 示例 1: 输入:head = [1,2,3,4,5], k = 2 输出:[2,1,4,3,5] 示例 2: 输入:head = [1,2,3,4,5], k = 3 输出:[3,2,1,4,5] 示例 3: 输入:head = [1,2,3,4,5], k = 1 输出:[1,2,3,4,5] 示例 4: 输入:head = [1], k = 1 输出:[1] (PS:还是加上title吧不然太难受了) 前言:LeetCode难度hard

Given a non-negative integer num represented as a string, remove k digits from the number so that the new number is the smallest possible. Note: The length of num is less than 10002 and will be ≥ k. The given num does not contain any leading zero. Ex

You are given two integer arrays nums1 and nums2 sorted in ascending order and an integer k. Define a pair (u,v) which consists of one element from the first array and one element from the second array. Find the k pairs (u1,v1),(u2,v2) ...(uk,vk) wit

Given a non-empty array of integers, return the k most frequent elements. For example,Given [1,1,1,2,2,3] and k = 2, return [1,2]. Note: You may assume k is always valid, 1 ≤ k ≤ number of unique elements.    Your algorithm's time complexity must be

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format. If the fractional part is repeating, enclose the repeating part in parentheses. For example, Given numerator = 1, denominator = 2, retu

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity. 这道题让我们合并k个有序链表,之前我们做过一道Merge Two Sorted Lists 混合插入有序链表,是混合插入两个有序链表.这道题增加了难度,变成合并k个有序链表了,但是不管合并几个,基本还是要两两合并.那么我们首先考虑的方法是能不能利用之前那道题的解法来解答此题.答案是肯定的,但是需要修改

A typical solution is heap based - "top K". Complexity is O(nlgk). typedef pair<int, unsigned> Rec; struct Comp { bool operator()(const Rec &r1, const Rec &r2) { return r1.second > r2.second; } }; class Solution { public: vector

https://oj.leetcode.com/problems/merge-k-sorted-lists/ 归并K已经整理阵列,和分析算法的复杂. 解决报告:无论是不考虑优化,最简单的实现是要重新走路List<ListNode>.对当中每一个链表同当前链表做一遍类似于归并排序最后一步的merge操作. 算法复杂度是O(KN) public class Solution { ListNode mergeTwoLists(ListNode list1, ListNode list2) { Lis

Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first. Example 1: Inpu

Given a sorted array, two integers k and x, find the k closest elements to x in the array. The result should also be sorted in ascending order. If there is a tie, the smaller elements are always preferred. Example 1: Input: [1,2,3,4,5], k=4, x=3 Outp

Given a string representing an expression of fraction addition and subtraction, you need to return the calculation result in string format. The final result should be irreducible fraction. If your final result is an integer, say 2, you need to change

Given a non-empty list of words, return the k most frequent elements. Your answer should be sorted by frequency from highest to lowest. If two words have the same frequency, then the word with the lower alphabetical order comes first. Example 1: Inpu

K Best Time Limit: 8000MS   Memory Limit: 65536K Total Submissions: 12812   Accepted: 3290 Case Time Limit: 2000MS   Special Judge Description Demy has n jewels. Each of her jewels has some value vi and weight wi. Since her husband John got broke aft

原题链接在这里:https://leetcode.com/problems/k-closest-points-to-origin/ 题目: We have a list of points on the plane.  Find the K closest points to the origin (0, 0). (Here, the distance between two points on a plane is the Euclidean distance.) You may return

There is a garden with N slots. In each slot, there is a flower. The N flowers will bloom one by one in N days. In each day, there will be exactly one flower blooming and it will be in the status of blooming since then. Given an array flowers consist

题目: 在仅包含 0 和 1 的数组 A 中,一次 K 位翻转包括选择一个长度为 K 的(连续)子数组,同时将子数组中的每个 0 更改为 1,而每个 1 更改为 0. 返回所需的 K 位翻转的次数,以便数组没有值为 0 的元素.如果不可能,返回 -1. 示例 1: 输入:A = [0,1,0], K = 1输出:2解释:先翻转 A[0],然后翻转 A[2].示例 2: 输入:A = [1,1,0], K = 2输出:-1解释:无论我们怎样翻转大小为 2 的子数组,我们都不能使数组变为 [1,1,

25. K 个一组翻转链表 给你一个链表,每 k 个节点一组进行翻转,请你返回翻转后的链表. k 是一个正整数,它的值小于或等于链表的长度. 如果节点总数不是 k 的整数倍,那么请将最后剩余的节点保持原有顺序. 示例 : 给定这个链表:1->2->3->4->5 当 k = 2 时,应当返回: 2->1->4->3->5 当 k = 3 时,应当返回: 3->2->1->4->5 说明 : 你的算法只能使用常数的额外空间. 你不能只是

题目链接 Merge k sorted 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,我们计算一下复杂度. 1.2合并,遍历2n个节点 12结果和3合并,遍历3n个节点 123

题目链接:http://poj.org/problem?id=3111 求选k对数,使得上述式子值最大.容易想到设左边为一个值,对式子变形以下,得到sigma(v-r*w))==0的时候就是最大的,<0是最小的.二分这个r就行了. #include <algorithm> #include <iostream> #include <iomanip> #include <cstring> #include <climits> #include

本题的特点在于两个list nums1和nums2都是已经排序好的.本题如果把所有的(i, j)组合都排序出来,再取其中最小的K个.其实靠后的很多组合根本用不到,所以效率较低,会导致算法超时.为了简便,把nums1和nums2里面的元素写成A1,A2,...,A5, B1,...,B5. 维护一个最小堆.并逐渐往里面push and pop元素. 注意以下几点:1. 如果(Ai, Bj)大于堆顶的元素,那么没有比较把(Ai, Bj)和它之后的元素Push进堆.2. 为了便于管理,解法里其实只维护