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.

解题思路：

这是关于分而治之算法的题目。运用递归的思想，将原来的数组减半分成两个数组，再计算出这两个数组的majority element。如果这两个数组的majority element相等的话，就说明这是原来数组的majority element，否则就比较这两个elements在原来数组中出现的次数，次数多的element才是原来数组真正的majority element。
一开始判断elements出现次数的时候，选择for循环遍历的粗暴方法：

 for (int i = begin; i <= end; i++) {
    if (nums[i] == left_element)
        num1++;
    else if (nums[i] == right_element)
        num2++;
 }

但是很明显，RunTime Error!!!之后发现STL算法中的count函数可以实现for循环的功能并且降低时间复杂度，试了一下就真的Accepted!!!

代码实现：

#include <vector.h>
#include <algorithm.h>

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int begin = 0;
        int end = nums.size()-1;    //注意“-1”
        return majority(nums, begin, end);
    }
 private:
    int majority(vector<int>& nums, int begin, int end) {
        //nums只有1个元素
        if (begin == end)
            return nums[begin];
        else {
            int middle = (end - begin)/2 + begin;   //注意begin,middle,end的大小关系
            int left_element = majority(nums, begin, middle);
            int right_element = majority(nums, middle+1, end);
            if (left_element == right_element)
                return left_element;
            else {
                int num1 = 0, num2 = 0;
                num1 = count(nums.begin()+begin, nums.begin()+end+1, left_element);
                num2 = count(nums.begin()+begin, nums.begin()+end+1, right_element);
                if (num1 >= num2) return left_element;
                else return right_element;
            }
        }
    }
};