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
(M) Majority Element II More readings:https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority_vote_algorithm publicclassSolution {publicintmajorityElement(int[] num) {intmajor = num[0];intcount = 1;//The majority number has enough counts to cover all other numbersfor(inti = 1; i <...
classSolution{publicintmajorityElement(int[]nums){intres=0;intcount=0;for(intnum:nums){if(count==0){res=num;}count+=(num==res)?1:-1;}returnres;} 有兴趣的话还可以再看一下下面这个 wiki上的伪代码: The algorithm maintainsinits local variables a sequence elementanda counter,withthe counter...
#include<iostream>#include<vector>#include<algorithm>intmajorityElement(std::vector<int>&nums){std::sort(nums.begin(),nums.end());returnnums[nums.size()/2];}intmain(){intarr[]={1,2,3,2,4,2,2,2,2,5,7};std::vector<int>nums(arr,arr+sizeof(arr)/sizeof(arr[0]));intresult=m...
229 Majority Element 2 这道题首先是受到 169 Majority Element 的启发 (https://en.wikipedia.org/wiki/Boyer-Moore_Majority_Vote_Algorithm) 知道最多有2个可能的数满足条件。 而这道题不同的是无法确定是否存在, 因此得到可能的candidates后需要重新扫描一遍数组 判定是否满足条件, 代码如下 时间 104ms...
https://leetcode.com/discuss/43248/boyer-moore-majority-vote-algorithm-and-my-elaboration http://stackoverflow.com/questions/24691048/find-all-elements-that-appear-more-than-n-4-times-in-linear-time https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority_vote_algorithm...
Algorithm For this algorithm, we simply do exactly what is described: sort nums, and return the element in question. To see why this will always return the majority element (given that the array has one), consider the figure below (the top example is for an odd-length array and the bott...
(https://en.wikipedia.org/wiki/Boyer-Moore_majority_vote_algorithm) 此方法需要O(n)的时间和O(1)的空间。首先将第一个数字假设为所求之数,将计数器设置为1。遍历数组,比较数组中的数和假设之数是否相等,若相等则计数器加1,否则减1。当计数器为0时,将当前遍历到的数字设为新的假设之数,计数器初始化为...
"""length =len(nums)/2listelement =set(nums)forxinlistelement:ifnums.count(x) > length:returnx 写的很粗糙,这个link比较分明 http://bookshadow.com/weblog/2014/12/22/leetcode-majority-element https://en.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority_vote_algorithm ...
Mehrheitselement finden (Boyer-Moore Majority Vote Algorithm)Gibt bei einem Integer-Array mit Duplikaten das Mehrheitselement zurück, falls vorhanden. Ein Mehrheitselement erscheint mehr als n/2 Mal, wo n ist die Arraygröße.Beispielsweise ist das Mehrheitselement 2 im Array {2, 8,...