面试题查找重复元素并打印重复次数和重复位置,一顿懵逼,回来死磕写下来,打印指定重复次数和最大次数,其他在此基础上可以再更新 package sort; import org.testng.annotations.Test;import sun.org.mozilla.javascript.internal.ast.NewExpression; import java.util.*; /** * Created by liangwei on 2018/10/18. */public class S…
Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times. Example 1: Input: s = "aaabb", k = 3 Output: 3 The longest substring is "aaa"…
Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times. Example 1: Input: s = "aaabb", k = 3 Output: 3 The longest substring is "aaa"…
前言 网站设计的优化是一个很大的话题,有一些通用的原则,也有针对不同开发平台的一些建议.这方面的研究一直没有停止过,我在不同的场合也分享过这样的话题. 作为通用的原则,雅虎的工程师团队曾经给出过35个最佳实践.这个列表请参考 Best Practices for Speeding Up Your Web Site http://developer.yahoo.com/performance/rules.html 同时,他们还发布了一个相应的测试工具Yslow http://developer.…
Once, Leha found in the left pocket an array consisting of n integers, and in the right pocket q queries of the form l r k. If there are queries, then they must be answered. Answer for the query is minimal x such that x occurs in the interval l r str…
阅读目录: DS01:常用的查找数组中是否有重复元素的三种方法 DS02:常用的JS函数集锦 DS01.常用的查找数组中是否有重复元素的三种方法 1. var ary = new Array("111","22","33","111"); var s = ary.join(",")+","; for(var i=0;i<ary.length;i++) { if(s.replace…
题目: Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. You may assume that the array is non-empty and the majority element always exist in the array. 思路:可以利用Dictionary将数组中每个数…
一.查找数列重复元素---count() >>> list = [,,,,,,,,,,,] >>> set = set(list) >>> for item in set: print("the %d has found %d" %(item,list.count(item))) #输出 #the has found #the has found #the has found #the has found 二.查找重复元素,使用 C…
题目: Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. 解析:同求全部组合的过程一样,只是这里限制每个组合中元素的个数为k,求所有组合时并不限制元素个数. 若要排除重复的元素,则首先对所有元素进行排序. 代码: public static List<List<Integer>> getAllCombinations(int[] array,int k)…
题目描述: Given an array of integers and an integer k, find out whether there are two distinct indices i and j in the array such that nums[i] = nums[j] and the absolute difference between i and j is at most k. Example 1: Input: [1,2,3,1], k = 3 Output: t…
题目原文: Decimal dominants. Given an array with n keys, design an algorithm to find all values that occur more than n/10 times. The expected running time of your algorithm should be linear. 分析: 直观上将n个元素遍历一遍,并记录每个元素出现的次数就可以实现,虽然时间复杂度是O(n),但是空间复杂度却高达n,这肯…
先讨论出现次数大于n/2的数字,如果这样的数字存在,那么这个数出现的次数大于其他数出现的次数的总和. 在数组A中,我们定义两个数据集合a1,a2.a1为出现次数大于n/2的数的集合,a2为其余数组成的集合.对于数组 A中元素a.b,假设a不等于b,那么有两种情况,分别为:a属于a1,b属于a2:a属于a2,b属于a2.对于这两种情况,如 果把a.b从数组A中去掉,集合a1的size依旧是大于a2的.按照这个思路,我们有如下代码: int m; ; for (auto num : nums) {…
Majority Number III 给定一个数组(长度为L),找到所有出现频次大于1/k的数字. 我们主要使用摩尔投票法(Voting Algorithm)结合Map的数据结构解决此问题.其时间复杂度O(n),空间复杂度O(k). 主元素数量大于数组长度的1/k,因此有k-1一个候选人(Candidate),通过loop整个数组两次得到答案,遍历第一次找有可能的候选人,第二次统计这些候选人出现的频次. 当数组长度L小于k时,直接对所有出现的数字进行统计得到结果. 当数组长度L大于等于k时,建…
Q: A: 分治,对于字符串s的任何一个字符,如果它的频数(在s中出现的次数)小于k,则它一定不会出现在最后的结果里,也就是从它的位置一劈两半,考察左右.对于当前字符串s,我们先建立字典统计其中每种字符出现的次数,对于某字符,假设为x,x在当前字符串中出现的次数小于为kk,kk<k.则所有的字符x可将当前字符串s切片为kk+1个子串,递归对这kk+1个子串进行考察即可. class Solution { public: int longestSubstring(string s, int k)…
用快速排序的思想输出数组第k大的元素: #include<iostream> #include<algorithm> using namespace std; //递归实现:返回数组第k大的值.数组下标区间是[begin,end].其中数组假定n个元素,则k的值在区间[1,n]. //能够使用这种方法的前提条件是:n个数不能重复.如果n个数中有重复,那么区间的大小不能保证就是第K大. int findkth(int* arr, int begin, int end, int k)…
