Majority Element II

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.

网络上的思路参考：

Find k different element, and "remove" them as a group, the remaining element must be the element that appears more than 

⌊n/k⌋ times. 至多有k-1个出现次数超过⌊ n/k ⌋的元素：至多有1个出现次数超过⌊ n/2 ⌋，至多有2个出现次数超过⌊ n/3 ⌋...

 

For those who aren't familiar with Boyer-Moore Majority Vote algorithm, I found a great article 

(http://goo.gl/64Nams) that helps me to understand this fantastic algorithm!! Please check it out!

The essential concepts is you keep a counter for the majority number X. If you find a number Ythat is not X, the current counter should deduce 1. The reason is that if there is 5 X and 4 Y, there would be one (5-4) more X than Y. This could be explained as "4 X being paired out by 4 Y".

And since the requirement is finding the majority for more than ceiling of [n/3], the answer would be less than or 

equal to two numbers. So we can modify the algorithm to maintain two counters for two majorities.

观察可知，数组中至多可能会有2个出现次数超过 ⌊ n/3 ⌋ 的众数

记变量n1, n2为候选众数； c1, c2为它们对应的出现次数

遍历数组，记当前数字为num

若num与n1或n2相同，则将其对应的出现次数加1

否则，若c1或c2为0，则将其置为1，对应的候选众数置为num

否则，将c1与c2分别减1

最后，再统计一次候选众数在数组中出现的次数，若满足要求，则返回之。

 

class Solution

{

public: 

     vector<int> majorityElement(vector<int> &a) 

 

     {

          int y = 0, z = 1, cy = 0, cz = 0;

          for (auto x: a) 

          {

               if (x == y) cy++;

               else if (x == z) cz++;

               else if (! cy) y = x, cy = 1;//仔细分析，确实很有道理，假设其中一个候选众数占了1/3（遇到众数，其 

他操作就可以忽略，请看！！！这些代码都是独立的，一次只能选择一条执行路径），那么另外一个候选众数（假如有）则占了 

剩下的元素的一半以上！！

               else if (! cz) z = x, cz = 1;

               else cy--, cz--;

          }

          cy = cz = 0; 

          for (auto x: a) if (x == y) cy++;

          else if (x == z) cz++;

          vector<int> r; 

          if (cy > a.size()/3) r.push_back(y); 

          if (cz > a.size()/3) r.push_back(z);

          return r;

      } 

};

来看我自己的实现代码：

#include<iostream>
#include<vector>
using namespace std;


class Solution {
public:
    vector<int> majorityElement(vector<int>& nums) {
        vector<int>res;
        int mode1=INT_MAX, mode2=INT_MIN,cn1=0,cn2=0;
        for(auto x:nums)
        {
            if (x == mode1)cn1++;
            else if (x == mode2)cn2++;
            else if (!cn1)mode1 = x, cn1 = 1;
            else if (!cn2)mode2 = x, cn2 = 1;
            else cn1--, cn2--;
        }
        cn1 = 0, cn2 = 0;
        for (auto x :nums)
        {
            if (x == mode1)cn1++;
            if (x == mode2)cn2++;
        }
        if (cn1 > nums.size() / 3)res.push_back(mode1);
        if (cn2 > nums.size() / 3)res.push_back(mode2);
        return res;
    }
};
int main()
{
    Solution test;
    vector<int> val = {3};//{ 2, 3, 5, 3, 2, 3,2,5};
    vector<int>result = test.majorityElement(val);
    for each(auto x in result)
        cout << x << " ";
    return 0;
}