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 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. 这一题可以用排序之后查看序列正中间那个元素的方法来解.但…

Leetcode之分治法专题-169. 求众数(Majority Element) 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. 示例 1: 输入: [3,2,3] 输出: 3 示例 2: 输入: [2,2,1,1,1,2,2] 输出: 2 分治法,顾名思义,分而治之,就是把要求解的问题,一分为二,在每个分支上再求解. 这题里,我们可以求出一个mid=(L+R)>>>1;求mid的…

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…

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-empty and the majority element always exist in the array. 思路1:ha…

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<…

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…

1. 题目描述Description Link: https://leetcode.com/problems/majority-element/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-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. 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. 分析: 遍历数组,每当发现一对儿不相同的element时…

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. 解题思路: 编程之美P130(寻找发帖水王)原题,如果删…

题目描述: 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. 解题思路: 每找出两个不同的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. 思路: 找到一个数组中出现次数超过一半的数.排序.哈希等…

题目要求 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. 题目分析及思路 给定一个长度为n的数组,找到m…

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. 也就是找数组中出现次数大于一半的数字,题目保证这…

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. Hide Tags: Divide and Conquer…

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…

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 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:设置两个变量,一个是current,初始化为A[0…

题目 A peak element is an element that is greater than its neighbors. Given an input array where num[i] ≠ num[i+1], find a peak element and return its index. The array may contain multiple peaks, in that case return the index to any one of the peaks is…

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-empty and the majority element always exist in the array. 找出数列中个…

LeetCode 81 Search in Rotated Sorted Array II [binary search] <c++> 给出排序好的一维有重复元素的数组,随机取一个位置断开,把前半部分接到后半部分后面,得到一个新数组,在新数组中查找给定数是否存在,时间复杂度限制\(O(log_2n)\) C++ 因为有重复元素存在,nums[l] <= nums[mid]不能说明[l,mid]区间内一定是单调的,比如数组[1,2,3,1,1,1,1],但是严格小于和严格大于的情况还是可以…

LeetCode 80 Remove Duplicates from Sorted Array II [Array/auto] <c++> 给出排序好的一维数组,如果一个元素重复出现的次数大于两次,删除多余的复制,返回删除后数组长度,要求不另开内存空间. C++ 献上自己丑陋无比的代码.相当于自己实现一个带计数器的unique函数 class Solution { public: int removeDuplicates(std::vector<int>& nums) {…

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(…

DOM操作时,经常使用element.style属性,没错,element.style是属性,和几个offsetXxxx属性一样,概念是一样的. 但是style有几个属性,这几个属性和offsetXxxx有很大关系.他们是可以相互转化的. 之所以说转化,是因为他们的值类型不同,element.offsetXxxx的值类型是Number,并且是整型,比如100.而element.style.xxx是带有单位的字符串,比如100px. 所以要相互转化,要使用parseInt对style的数据进行处理…

<divid="test1"> <a href="http://www.hujuntao.com/">设计蜂巢</a> </div> <pid="bar">111</p> <script> var d1 = document.getElementById('test1'), obj1 = d1.querySelector('div a'), obj2 = d1.q…

Leetcode之二分法专题-275. H指数 II(H-Index II) 给定一位研究者论文被引用次数的数组(被引用次数是非负整数),数组已经按照升序排列.编写一个方法,计算出研究者的 h 指数. h 指数的定义: “h 代表“高引用次数”(high citations),一名科研人员的 h 指数是指他(她)的 (N 篇论文中)至多有 h 篇论文分别被引用了至少 h 次.(其余的 N - h 篇论文每篇被引用次数不多于 h 次.)" 示例: 输入: citations = [0,1,3,5,…

Leetcode之回溯法专题-90. 子集 II(Subsets II) 给定一个可能包含重复元素的整数数组 nums,返回该数组所有可能的子集(幂集). 说明:解集不能包含重复的子集. 示例: 输入: [1,2,2] 输出: [ [2], [1], [1,2,2], [2,2], [1,2], [] ] 分析:是78题的升级版,新增了一些限制条件,nums数组是会重复的,求子集. class Solution { List<List<Integer>> ans = new Arr…

Leetcode之回溯法专题-47. 全排列 II(Permutations II) 给定一个可包含重复数字的序列,返回所有不重复的全排列. 示例: 输入: [1,1,2] 输出: [ [1,1,2], [1,2,1], [2,1,1] ] 分析:跟46题一样,只不过加了一些限制(包含了重复数字). AC代码(时间复杂度比较高,日后二刷的时候改进): class Solution { List<List<Integer>> ans = new ArrayList<>()…

目录 # 前端与算法 leetcode 350. 两个数组的交集 II 题目描述 概要 提示 解析 解法一:哈希表 解法二:双指针 解法三:暴力法 算法 # 前端与算法 leetcode 350. 两个数组的交集 II 题目描述 给定两个数组,编写一个函数来计算它们的交集. 示例 1: 输入: nums1 = [1,2,2,1], nums2 = [2,2] 输出: [2,2] 示例 2: 输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4] 输出: [4,9] 说明…