Leetcode 229 Majority Element II

The basic algorithm used in this problem is Boyer–moore majority vote algorithm

There are sample code in the Wiki

 1 Import java.util.*; 2 public class Majorityvote {3 public int majorityelement (int[] num) {4 int n = num.length; 5 int CA Ndidate = num[0], counter = 0;  6 for (int i:num) {7 if (counter = = 0) {8 candidate = i; 9 counter = 1;10} else {one (i = = candidate) {counter++;13}         else {counter--;15}16}17}18 counter = 0;20 for (int i:num) {if (i = = candidate) counter++;22}23 if (counter < (n + 1)/2) RET urn-1;24 return candidate;25}27 public static void Main (string[] args) {Majorityvote s = n EW Majorityvote (); System.out.format ("%d\n", S.majorityelement (new int[] {1, 2, 3})); T ("%d\n", S.majorityelement (new int[] {2, 2, 3}); 31}32}

The basic idea is this: there is at most one element in the array that has more than N/2, so there is only one candidate element during the traversal

We assume that there is an element that satisfies this requirement: if he is evenly distributed, at least once every interval, and not enough, at least one more time at some point.

Now think about the uneven situation: if the element is spaced out for a long time, there is at least a dense area in front or behind the long interval.

Vector<int> majorityelement (vector<int>& nums) {if (Nums.empty ()) return vector<int> ();    if (nums.size () = = 1) {return vector<int> ({nums[0]});    } vector<pair<int, int>> candidates;        for (auto num:nums) {if (Candidates.size () < 1) {candidates.emplace_back (num, 2); } else if (Candidates.size () < 2) {if (num! = Candidates[0].first) Candidates.emplace_ba            CK (num, 2);        else candidates[0].second++;                } else{if (num = = Candidates[0].first) {candidates[0].second++;            candidates[1].second--;                } else if (num = = Candidates[1].first) {candidates[1].second++;            candidates[0].second--;                } else{candidates[0].second--;                candidates[1].second--; int ind = Candidates[0].second < candidates[1]. Second?                0:1;                    if (candidates[ind].second <= 0) {candidates[ind].first = num;                Candidates[ind].second = 2;    }}}} for (auto& candidate:candidates) {candidate.second = 0;                } for (auto Num:nums) {for (auto& candidate:candidates) {if (num = = Candidate.first)        candidate.second++;    }} vector<int> result; for (auto candidate:candidates) {if (Candidate.second > Nums.size ()/3) Result.push_back (Candidate.fir    ST); } return result;

　　

Leetcode 229 Majority Element II