题目：
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.
题解：
将数组二分，然后进行递归，递归到一就是当前的众数，再合并，比较两个众数取次数多的往上并，中间用到计数函数count。
代码：
class Solution { public: int majorityElement(vector<int>& nums) { return majority(nums, 0, nums.size() - 1); } private: int majority(vector<int>& nums, int left, int right) { if (left == right) return nums[left]; int mid = left + ((right - left) >> 1); int lm = majority(nums, left, mid); int rm = majority(nums, mid + 1, right); if (lm == rm) return lm; return count(nums.begin() + left, nums.begin() + right + 1, lm) > count(nums.begin() + left, nums.begin() + right + 1, rm) ? lm : rm; } };
