[LeetCode]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.

这道题是计算出一组数中出现频率大于? n/2 ?次的那个数。我的思路简单粗暴，使用STL中的map，键为数组中的元素，值为元素对应的频率。遍历一边数组，计算各自的频率，最后选出map中值大于? n/2 ?对应的键。

贴上代码：

class Solution {
public:
    int majorityElement(vector<int> &num) {
        map<int, int> m;
        for (int i = 0; i < num.size(); i++){
            m[num[i]]++;
        }
        map<int, int>::iterator it = m.begin();
        int count = num.size() / 2;
        for (; it != m.end(); it++){
            if (it->second >= count)
                return it->first;
        }
    }
};

当然了，简单粗暴的结果就是74ms…QAQ
算了，不高兴优化了【看到有人一行Python就解决了 o(>﹏<)o】