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.
思路：
找主要元素，用major记录主要字母，n记录major相对于其他字母多出现的次数
if n == 0 ， 设当前数字为主要数字
else if 当前数字等于主要数字， n++
else 当前数字不等于主要数字， n--
因为主要数字出现多于一半，所以最后major表示的一定是主要数字。
没测试就一次AC了
class Solution { public: int majorityElement(vector<int> &num) { int major; int n = 0; for(int i = 0; i < num.size(); i++) { if(n == 0) { major = num[i]; n++; } else if(major == num[i]) { n++; } else { n--; } } return major; } };
【leetcode】Majority Element (easy)(*^__^*)
原文：http://www.cnblogs.com/dplearning/p/4214245.html