169. Majority Element /** * @param {number[]} nums * @return {number} */ var majorityElement = function(nums) { var hash = {}; var y=-1,z; //注意这里的方括号,利用变量访问对象属性时要用方括号 for(var i=0;i<=nums.length-1;i++){ if(hash[nums[i]]){ hash[nums[i]]++; }else{ hash[…

#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…

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 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(…

题目描述 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 思路 思路一: 利用哈希表的映射,储存数组中的数字以及它们出现的次数,当众数出现时,返回这个数字. 思路二: 因为众数是出现次数大于n/2的数字,所以排序之后中间的那个数字一定是众数.即nums[n/2]为众数.但是在计算比…

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…

Question 169. Majority Element Solution 思路:构造一个map存储每个数字出现的次数,然后遍历map返回出现次数大于数组一半的数字. 还有一种思路是:对这个数组排序,次数超过n/2的元素必然在中间. Java实现: public int majorityElement(int[] nums) { Map<Integer, Integer> countMap = new HashMap<>(); for (int num : nums) { In…

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…

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. 题目大意:…

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<…