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。