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.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

每找出两个不同的element，则成对删除。最终剩下的一定就是所求的。

(因为比别人多还有另外一种解释，就是你的票抵消了我的一部分票后，我还有剩。)

可扩展到? n/k ?的情况，每k个不同的element进行成对删除。

class Solution {
public:
    int majorityElement(vector<int> &num) {
        int nTimes = 0;
        int candidate = 0;
        for(int i = 0; i < num.size(); i ++)
        {
            if(nTimes == 0)//选择新的候选
            {
                candidate = num[i];
                nTimes = 1;
            }
            else
            {
                if(candidate == num[i])
                    nTimes ++;
                else
                    nTimes --;
            }
        }
        return candidate;
    }
};

Majority Element

原文：http://blog.csdn.net/sxhlovehmm/article/details/46433199