169. Majority Element My Submissions Question

Total Accepted: 95925 Total Submissions: 239241 Difficulty: Easy

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.

1. O(n*n)：判断每个元素是否是majority

2. O(n): 利用HashTable存储每个元素的个数，空间复杂度高

3. O(nlogn)：对数组排序，majority肯定在n/2位置处。

4. O(nlogn)：分而治之，分成两分A和B，如果A和B的majority相等，则其为整个数组的majority；否则，分别扫描计算A，B的majority元素个数(O(2n))，所以时间复杂度T(n) = T(n/2) + 2n = O(nlogn)。

5. O(n)：两两删除数组中两个不同的元素，最后剩下的就是多的元素。代码如下：

    int majorityElement(vector<int>& nums) {
        if (nums.empty())
            return -1;
   
        int candidate, count = 0;
   
        for (int i = 0; i < nums.size(); ++i) {
            if (count == 0) {
                candidate = nums[i];
                ++count;
            }
            else {
                if (candidate == nums[i])
                    ++count;
                else
                    --count;
            }
        }
   
        return candidate;
    }