Majority Element
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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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