Given an array of sizen, 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 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. 方法一: 采用所谓的Moore voting algorithm: 每找出两个不同的element,就成对删除即count–,最终剩下的一定就是所求的...
这题比较有意思的是,有一个优于hash map的算法 - 由Robert S. Boyer和J Strother Moore发表的MJRTY - A Fast Majority Vote Algorithm。 这个算法有一个限制就是这个众数必须要出现大于⌊ n/2 ⌋次。 由可以看到wikipedia伪代码可以写成: Initialize an elementm and a counteri withi= 0 For each ele...
1,定义Majority的候选candidate和计数cnt,并初始化为0。从i=0开始,顺序遍历; 2,if(cnt==0),令candidate = num[i], cnt = 1; 3,否则,if(num[i]==candidate) cnt++;否则,cnt--; 4,遍历完成,返回candidate即可。 看到该解法的第一眼,一脸懵逼,甚至怀疑代码写错了。直到看懂Algorithm的详细解释,才恍然大...
The following passage will implement 6 of them except the O(n^2) brute force algorithm. Hash Table The hash-table solution is very straightforward. We maintain a mapping from each element to its number of appearances. While constructing the mapping, we update the majority element based on the...
// Majority Element// Time Complexity: O(nlogn), Space Complexity: O(1)classSolution{public:intmajorityElement(vector<int>&nums){sort(nums.begin(),nums.end());returnnums[nums.size()/2];}}; 解法2 Boyer-Moore Voting Algorithm # Majority Element# Time Complexity: O(nlogn), Space Complexi...
/majority-element-ii/description/ 题目描述: 知识点: 思路一:用哈希表记录各个数字出现的次数 时间复杂度和空间复杂度均是O(n),其中n为数组的长度。 JAVA代码: LeetCode解题报告: 思路二:Boyer–Moore majority vote algorithm 题目要找的数要求出现次数大于n / 3(向下取整)次,因此不会多于2个数,即满足条件...
主元素问题是一个非常经典的问题,一般来说,主元素问题指的是数组中元素个数大于一半的数字,显然这个问题可以通过遍历计数解决,时间复杂度为O(n),空间复杂度为O(n)。这样的算法有两个弊端,一是空间复杂度较高,二是没法处理数据流问题。 因此就有了Boyer-Moore Majority Vote algorithm,这个算法可以用来高效的解决...
229. Majority Element II #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 这道题目求频繁项,且数字次数大于 ⌊ n/3 ⌋ ,这个限制条件就限制了频繁项的个数一定小于...
Description 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: ...