题目描述: Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 分析: 因为要找出的是出现次数大于⌊ n/3 ⌋的元素,因此最多只可能存在两个这样的元素,而且要求O(1)的空间复杂度,因此只能使用摩尔投票法.首先我们遍历一遍数组找出两个候选元素,接着再遍历…

先看一题,洛谷2397: 题目背景 自动上次redbag用加法好好的刁难过了yyy同学以后,yyy十分愤怒.他还击给了redbag一题,但是这题他惊讶的发现自己居然也不会,所以只好找你 题目描述 [h1]udp2:第一题因为语言性质问题,比赛结束后将所有c/c++的程序的内存调为2.2mb后重测.[/h1] 他让redbag找众数 他还特意表示,这个众数出现次数超过了一半 一共n个数,而且保证有 n<=2000000 而且每个数<2^31-1 代码 #include<bits/stdc+…

就是简单的应用多数投票算法(Boyer–Moore majority vote algorithm),参见这道题的题解. class Solution { public: vector<int> majorityElement(vector<int>& nums) { ,cnt2=,ans1=,ans2=; for(auto n:nums){ if(n==ans1){ cnt1++; } else if(n==ans2){ cnt2++; } ){ ans1=n; cnt1…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. Note: The algorithm should run in linear time and in O(1) space. Example 1: Input: [3,2,3] Output: [3] Example 2: Input: [1,1,1,3,3,2,2,2] Output: [1,2] 169. Maj…

这题用到的基本算法是Boyer–Moore majority vote algorithm wiki里有示例代码 1 import java.util.*; 2 public class MajorityVote { 3 public int majorityElement(int[] num) { 4 int n = num.length; 5 int candidate = num[0], counter = 0; 6 for (int i : num) { 7 if (counter ==…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 题目标签:Array 题目给了我们一个 nums array, 让我们找出所有 数量超过 n/3 的众数.这一题与 众数之一 的区别在于,众数之一 只要找到一个 众数大于 n/2 的就可以.这一题要找到所…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 解题思路: <编程之美>寻找发帖水王的原题,两次遍历,一次遍历查找可能出现次数超过nums.length/3的数字,(出现三次不同的数即抛弃),第二次验证即可. JAVA实现如下: public Lis…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 解法:参考编程之美129页,寻找发帖水王 代码如下: public class Solution { public List<Integer> majorityElement(int[] nums) {…

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. Example 1: Input: [3,2,3] Ou…

给定一个大小为 n 的数组,找出其中所有出现超过 ⌊ n/3 ⌋ 次的元素. 你的算法应该在O(1)空间中以线性时间运行. 详见:https://leetcode.com/problems/majority-element-ii/description/ 摩尔投票法 Moore Voting Java实现: class Solution { public List<Integer> majorityElement(int[] nums) { List<Integer> res=ne…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. Hint: How many majority elements could it possibly have? Do you have a better hint? Suggest it! 这道题让我们…

169. Majority Element 求超过数组个数一半的数 可以使用hash解决,时间复杂度为O(n),但空间复杂度也为O(n) class Solution { public: int majorityElement(vector<int>& nums) { unordered_map<int,int> count; int n=nums.size(); ;i<n;i++){ ) return nums[i]; } ; } }; 使用投票法,时间复杂度为O(…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. Note: The algorithm should run in linear time and in O(1) space. Example 1: Input: [3,2,3] Output: [3] Example 2: Input: [1,1,1,3,3,2,2,2] Output: [1,2] 这道题让我们求出…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 hashmap统计次数 摩尔投票法 Moore Voting 相似题目 参考资料 日期 题目地址:https://leetcode.com/problems/majority-element-ii/description/ 题目描述 Given an integer array of size n, find all elements that ap…

Majority Element II Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. Note: The algorithm should run in linear time and in O(1) space. Example 1: Input: [3,2,3] Output: [3] Example 2: Input: [1,1,1,3,3,2,2,2] Ou…

一个数组里有一个数重复了n/2多次,找到 思路:既然这个数重复了一半以上的长度,那么排序后,必然占据了 a[n/2]这个位置. class Solution { public: int majorityElement(vector<int>& nums) { sort(nums.begin(),nums.end()); return nums[nums.size()/2]; } }; 线性解法:投票算法,多的票抵消了其余人的票,那么我的票一定还有剩的. int majority; in…

Majority Element II Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. Hint: How many majority elements could it possibly have? Do you have a better hint…

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. 题目标签:Array 忘记说了,特地回来补充,今天看完<…

题目: Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 思路: 首先,我们来看一下怎样求众数,也就是元素出现大于⌊ n/2 ⌋的数. 我们注意到这样一个现象: 在任何数组中,出现次数大于该数组长度一半的值只能有一个. 通过数学知识,我们可以证明它的正确…

寻找多数元素这一问题主要运用了:Majority Vote Alogrithm(最大投票算法)1.Majority Element 1)description 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…

Boyer-Moore majority vote algorithm(摩尔投票算法) 简介 Boyer-Moore majority vote algorithm(摩尔投票算法)是一种在线性时间O(n)和空间复杂度的情况下,在一个元素序列中查找包含最多的元素.它是以Robert S.Boyer和J Strother Moore命名的,1981年发明的,是一种典型的流算法(streaming algorithm). 在它最简单的形式就是,查找最多的元素,也就是在输入中重复出现超过一半以上(n/2…

原题网址; https://www.lintcode.com/problem/majority-element-ii/ 描述 给定一个整型数组,找到主元素,它在数组中的出现次数严格大于数组元素个数的三分之一. 数组中只有唯一的主元素 您在真实的面试中是否遇到过这个题?  是 样例 给出数组[1,2,1,2,1,3,3] 返回 1 挑战 要求时间复杂度为O(n),空间复杂度为O(1). 查看标签 枚举法 贪心   思路:利用map,建立nums[i]与数量count 的映射,最后返回 count…

题目 来源:力扣(LeetCode) 链接:https://leetcode-cn.com/problems/majority-element/ 注意,该题在LC中被标注为easy,所以我们更多应该关注的是文章中不断优化的思路和方法.很多时候面试考察的,就是与面试官一起做题并把时间复杂度和空间复杂度压榨到极致的过程中你所表现出来的能力. 1.描述 给定一个大小为 n 的数组,找到其中的多数元素.多数元素是指在数组中出现次数大于 ⌊n/2⌋的元素. 你可以假设数组是非空的,并且给定的数组总是存在多…

一:Majority Element 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. class Sol…

题目:找出数组中出现次数大于n/3次的数字 思路:摩尔投票法.所有的帖子中都说:先遍历一遍数组找到备选元素,然后再遍历一遍数组考察下这个元素是否是真的超过n/3,然后就直接上代码,但是现在的问题是:我怎么找到这个备选的元素?!票是怎么投起来的呢?通过参考文章中的代码,大致明白了,醍醐灌顶(http://www.cnblogs.com/grandyang/p/4606822.html).那么怎么遍历一遍数组然后找到备选元素呢?这个算法还得从找到大于一半的数说起:整个数组中有两个不一样的数的时候,就…

LeetCode169. Majority Element Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. (Easy) You may assume that the array is non-empty and the majority element always exist in th…

剑指 Offer 39. 数组中出现次数超过一半的数字 Offer_39 题目描述 方法一:使用map存储数字出现的次数 public class Offer_39 { public int majorityElement(int[] nums) { Map<Integer,Integer> map = new HashMap<>(); int len = nums.length; int ans = -1; for(int num : nums){ int cnt = 0; if…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. 思路: [LeetCode 169]Majority Element 的拓展,这回要求的是出现次数超过三分之一次的数字咯,动动我们的大脑思考下,这样的数最多会存在几个呢,当然是2个嘛.因此,接着上一题的方…

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. 题解:运用多数投票算法的思路来解:从头到尾遍历数…

Given an integer array of size n, find all elements that appear more than ⌊ n/3 ⌋ times. The algorithm should run in linear time and in O(1) space. Hint: How many majority elements could it possibly have? Do you have a better hint? Suggest it! 参考Lint…