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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

直接上代码吧

class Solution {
public:
    int majorityElement(vector<int>& nums) 
    {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        int cnt = 0;
        int res = 0;
        for (int i = 0; i < n; ++i)
        {
            int s = count(nums.begin(), nums.end(), nums[i]);
            if (cnt < s)
            {
                cnt = s;
                res = nums[i];
            }
            i += s;
        }    
        return res;
    }
};    

意思应该比较明了，就是不是很快速，也不简洁，看看其他大神的解答吧

这个哈希表应该能想到的呀~~~刷题比较少没有感觉~~~

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        unordered_map<int, int> counts; 
        int n = nums.size();
        for (int i = 0; i < n; i++)
            if (++counts[nums[i]] > n / 2)
                return nums[i];
    }
};

用到了nth_element函数，既然是多于一半的数自然中间.

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

既然用了上面这个函数，那么我为什么不直接sort就行了=_+, 真的太笨了。

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

先记这些吧~~~