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. 需要注意的是:这道题只是要找出多数元素,已经默认存在多数…

Reverse Linked List I Question Solution Reverse a singly linked list. Reverse Linked List I 设置三个指针即可,非常简单: /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ cla…

There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)). 看到这道题的通过率很诧异,感觉这道题挺容易的,因为其实它的思想还是很简单的. 1)最笨的方法去实现,利用排序将两个数组合并成一个数组,然后返回中位数,这种方…

题外话 最近有些网友来信问我博客怎么不更新了,是不是不刷题了,真是惭愧啊,题还是在刷的,不过刷题的频率没以前高了,看完<算法导论>后感觉网上很多讨论的题目其实在导论中都已经有非常好的算法以及数学证明,只是照搬的话好像意义也不是很大,希望找到些有代表性的题目在更新,另外希望能接着前面的<穷举递归和回溯算法终结篇>一系列如动态规划.贪心算法类的终结篇,在梳理自己知识结构的同时也能够帮助读者们更系统的学习算法思想.好了话不多说,进入正题. 问题描述 给定一个数组A[n], 定义数组的主元…

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…

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 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. 给定一个数组,求其中权制最大的元素,(该元素出现超过了一…

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. 个人博客:http://www.cnblogs.com/…

题目来源: 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…

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. 代码:oj测试通过 Runtime: 197 m…

这周刚开始讲了一点Divide-and-Conquer的算法,于是这周的作业就选择在LeetCode上找分治法相关的题目来做. 169.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-e…

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 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]…

一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)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…

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: Find k different 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. Example 1: Input: [3,2,3] Ou…

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…

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…

problem:Given an array of size n, find the majority element. The majority element is the element that appears more than ⌊ n/2 ⌋ times. problem analysis: 1.首先,需要统计数组元素出现的次数,包括不同的元素有几个?每一个不同的元素出现了几次?同时需要将不同的元素及出现的频率正确对应. 2.在第一步完成之后,可以寻找最大的频率数,然后找到major…

一道Matlab编程题 & 暴力解法 Matlab课上老师出了这样一道题: 一个篮子有K个鸡蛋: 2个2个拿剩1个: 3个3个全部拿完: 4个4个拿剩1: 5个5个拿剩4个: 6个6个拿剩3个: 7个7个拿全部拿完: 8个8个拿剩1个: 9个9个拿全部拿完: 求篮子里鸡蛋的个数K 虽然这是一道matlab拿来玩的题目,可是我觉得完全可以拿来做笔试题或者面试题.仔细想还是有点考算法能力的. 这道题直观地想是非常简单的,简单一想就可以发现鸡蛋个数一定是7,9,3的最小公倍数63的N倍,然后我们就可以…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4048 访问. 给定一个大小为 n 的数组,找到其中的众数.众数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众数. 输入: [3,2,3] 输出: 3 输入: [2,2,1,1,1,2,2] 输出: 2 Given an array of size n, find the majority element. T…

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 思路 hashmap统计次数 摩尔投票法 Moore Voting 位运算统计位数 相似题目 参考资料 日期 题目地址:https://leetcode.com/problems/majority-element/ Total Accepted: 110538 Total Submissions: 268289 Difficulty: Easy 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. Hint: How many majority elements could it possibly have? Do you have a better hint? Suggest it! 这道题让我们…

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+…

#Method 1import math class Solution(object):    def majorityElement(self, nums):        numsDic={}        for num in nums:            numsDic[num]=numsDic[nums]+1 if num in numsDic else 0            if numsDic[num]>len(nums)/2:                return…

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…

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 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: 找到一个数组中出现次数超过一半的数.排序.哈希等…

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…

题目地址: https://oj.leetcode.com/problems/majority-element/ 题目内容: 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…