【Leet Code】229. Majority Element II---Medium

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?Show More Hint

思路1：

题目中重点强调出现的次数大于⌊ n/3 ⌋，所以可能有0个、1个或者2个（最多2个）这样的数存在，所以只需要设置2个变量cand1,、cand2来记录出现次数可能大于⌊ n/3 ⌋的数据，分别用count1和count2记录数据出现的次数。

思路2：

另一种最直接的方法就是用map对数组中每个值出现的次数做记录，然后把出现次数大于⌊ n/3 ⌋的值存入返回结果，该方法的缺点是空间复杂度为O(n)。

代码1实现：

class Solution {
public:vector<int> majorityElement(vector<int>& nums) {vector<int> result;if(nums.size() < 1) return result;if(nums.size() == 1) return nums;int cand1 = 0, cand2 = 0;int count1 = 0, count2 = 0;//找到满足条件的数，可能有一个满足条件的，最多有两个满足条件的for(auto num: nums){if (count1 == 0)cand1 = num;else if (count2 == 0)cand2 = num;//处理count的值if(cand1 == num)++count1;else if(cand2 == num)++count2;else{--count1;--count2;}}if(count(nums.begin(), nums.end(), cand1) > nums.size() / 3)result.push_back(cand1);//此处cand1 != cand2一定要判断，否则对于数组[2,2]，就会返回[2,2]，而期望的结果是[2]if(cand1 != cand2 && count(nums.begin(), nums.end(), cand2) > nums.size() / 3)result.push_back(cand2);return result;}

};

代码实现2：

class Solution {
public:vector<int> majorityElement(vector<int>& nums) {map<int, int> myMap;for (auto& num: nums) myMap[num]++;vector<int> res;for (auto it = myMap.begin(); it != myMap.end(); it++) if (it->second > nums.size()/3)res.push_back((*it).first);return res;}
};

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