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…
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…
网址:https://leetcode.com/problems/majority-element-ii/ 参考:https://blog.csdn.net/u014248127/article/details/79230221 摩尔投票算法( Boyer-Moore Voting Algorithm) class Solution { public: vector<int> majorityElement(vector<int>& nums) { vector<in…
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…
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 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: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+…