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…

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…

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…

原题链接在这里:Majority Element I,Majority Element II 对于Majority Element I 来说,有多重解法. Method 1:最容易想到的就是用HashMap 计数,数值大于n/2(注意不是大于等于而是大于),就是返回值.Time O(n), Space O(n). Method 2: 用了sort,返回sort后array的中值即可.Time O(n*logn), Space O(1). Method 3: 维护个最常出现值,遇到相同count+…

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(…

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…

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…

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. Credits:Special thanks to @t…

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…

原题链接在这里:https://leetcode.com/problems/majority-element/ 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 ele…

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…

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…

Majority Element 本题收获: 1.初步了解hash,nth_element的用法 2.题目的常规思路 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…

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…

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…

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…

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…

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! [题目分析]…

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. class Solution { public: //O(n) Space compexity vector<int> majorityElement01(vector<int>&…

Total Accepted: 23103 Total Submissions: 91679 Difficulty: Medium 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 cou…

在一个数组中找到主要的元素,也就是出现次数大于数组长度一半的元素.容易想到的方式就是计数,出现次数最多的就是majority element,其次就是排序,中间的就是majority element.但是还有两种更有意思的实现方式时间效率O(n),空间效率O(1):1.Moore voting algorithm 投票算法,因为符合要求的majority element总是存在的,所以首先置计数器count=1,并选择数组的第一个元素作为candidate,往后遍历并计数,与candidate相…

Question 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. Majority Element I A Linear Time Voting Algorithm Solution 1 -- Binary Search 我们发现一个规律.如果是大于…

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 Solution: # @param num…

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…

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. 题目大意:给定一个大小为n的数组,找出Majority元素,满足出现次数大于 ⌊ n/3 ⌋ 次. 解题思路:大于 ⌊ n/3 ⌋ 次,那么应该最多有两个Majority元素,之前有道题是找出 大于⌊ n…

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…

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. 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…