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.

题目大意：

找出数组中超过一半的数。

 

C++实现代码：

#include<iostream>

#include<vector>

using namespace std;



class Solution {

public:

    int majorityElement(vector<int> &num) {

        if(num.empty())

            return -1;

        int n=num.size();

        int i;

        int index=0;

        int count=1;

        for(i=1;i<n;i++)

        {

            if(num[i]==num[index])

            {

                count++;

            }

            else

                count--;

            if(count<0)

            {

                count=1;

                index=i;

            }

        }

        return num[index];

    }

};



int main()

{

    vector<int> num={3,3,3,3,3,3,1,2,4,5,3,45,2,54};

    Solution s;

    cout<<s.majorityElement(num)<<endl;

}

 

#include<iostream>

#include<vector>

using namespace std;



class Solution {

public:

    int majorityElement(vector<int> &num) {

        if(num.empty())

            return -1;

        int n=num.size();

        int major=num[0];

        int i;

        int count=1;

        for(i=1;i<n;i++)

        {

            if(num[i]==major)

                count++;

            else

                count--;

            if(count<0)

            {

                count=1;

                major=num[i];

            }

        }

        return major;

    }

};



int main()

{

    vector<int> num={3,3,3,3,3,3,1,2,5,3,45,2,54};

    Solution s;

    cout<<s.majorityElement(num)<<endl;

}

 如果要找当好出现一半的数呢？此时这个数可能是通过上面的办法找到的那个数，也可能是最后一个数，因此，只需要再次遍历数组，找出与最后一个数相等的数的个数，如果小于一半，那么上面找出的数就是刚好出现一半的数，否则最后一个数是刚好出现一半的数。