Majority Element

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.

给定size 为n的数组，查找出主元素，就是出现次数大于n/2次的元素。你可以假定数组非空，而且主元素一定存在。

class Solution {
public:int majorityElement(vector<int>& nums) {map<int, int> im;for (int i = 0; i < nums.size(); ++i){map<int, int>::iterator it = im.find(nums[i]);if (it == im.end()) {im[nums[i]] = 1;} else {im[nums[i]]++;}if (im[nums[i]] > nums.size()/2) {return nums[i];}}return 0;}
};

有一种算法叫 Moore’s Voting Algorithm，由Robert S.Boyer 和J Strother Moore于1980年发明，是线性时间复杂度。

int majorityElement(vector<int> &num) {int majority;int cnt = 0;for(int i=0; i<num.size(); i++){if ( cnt ==0 ){majority = num[i];cnt++;}else{majority == num[i] ? cnt++ : cnt --;if (cnt >= num.size()/2+1) return majority;}}return majority;
}

当然，这种算法对于存在主元素的数组是有效的，如：

A A A C C B B C C C B C C

它肯定能返回主元素C。但是，如果不存在主元素，那么得到的结果就跟遍历顺序有关了。如:

A A A C C C B

如果是从左到右，那么结果是B，如果是从右到左，那么结果是A。

class Solution {
public:int majorityElement(vector<int>& nums) {int n = nums.size();  sort(nums.begin(),nums.end());  return nums[n/2]; }
};

#include<iostream>
#include<vector>
#include<algorithm>
#include<map>
using namespace std;
int major(vector<int>);
int main()
{vector<int>a = { 1, 2, 3, 4, 2, 2, 5, 2, 2, 2, 2, 2, 2 };cout << major(a) << endl;system("pause");return 0;}
int major(vector<int> a)
{/*sort(a.begin(), a.end());return a[a.size() / 2];*//*map<int, int> m;for (int i = 0; i < a.size(); i++){if (m.find(a[i]) == m.end()){m[a[i]] = 1;}else{m[a[i]]++;}if (m[a[i]]>a.size() / 2)return a[i];}*///标志位想减法 可以判断出现次数做多数int major;int num = 0;for (int i = 0; i < a.size(); i++){if (num == 0){major = a[i];num++;}else{if (major != a[i])num--;else{num++;}}if (num > a.size() / 2) return major;}}

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