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
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) {
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
import java.util.*; public class MaxRandom { public static void main(String[] args){ int[] num = new int[10]; for(int i=0;i<10;i++){ num[i]=getRandomNum(); System.out.print(num[i]+" "); } System.out.println(); FindMax(num); } public static in
Life Forms Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 13063 Accepted: 3670 Description You may have wondered why most extraterrestrial life forms resemble humans, differing by superficial traits such as height, colour, wrinkles, e
Super Mario Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3625 Accepted Submission(s): 1660 Problem Description Mario is world-famous plumber. His “burly” figure and amazing jumping ability
总结:数组内的数据要循环,外面的次数也要循环 package com.aini; import java.util.Scanner; //要求可以控制输入的次数,比如4次.则输入5个数,5组输出结果 public class DFSDFG { public static void main(String[] args) { Scanner c = new Scanner(System.in); int x = c.nextInt(); int a[] = new int[x]; int s =
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