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.

直接上代碼吧

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int>& nums) 
 4     {
 5         int n = nums.size();
 6         sort(nums.begin(), nums.end());
 7         int cnt = 0;
 8         int res = 0;
 9         for (int i = 0; i < n; ++i)
10         {
11             int s = count(nums.begin(), nums.end(), nums[i]);
 
12             if (cnt < s)
13             {
14                 cnt = s;
15                 res = nums[i];
16             }
17             i += s;
18         }    
19         return res;
20     }
21 };    

意思應該比較明了，就是不是很快速，也不簡潔，看看其他大神的解答吧

這個哈希表應該能想到的呀~~~刷題比較少沒有感覺~~~

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函數，既然是多於一半的數自然中間.

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

既然用了上面這個函數，那麽我為什麽不直接sort就行了=_+, 真的太笨了。

 1 class Solution {
 2 public:
 3     int majorityElement(vector<int>& nums) 
 4     {
 5         int n = nums.size();
 6         
 7         sort(nums.begin(), nums.end());
 8         
 9         return nums[n/2];
10     }
11 };

先記這些吧~~~

Majority Element