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.

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

Subscribe to see which companies asked this question

这道题我还是用map做的，对于某个关键字有这种计数类的性质的题目用map真的很好用，关键点还是记得map里面的元素是pair，我们需要（*i）.first，最后是32ms，还不错吧

一次提交就accepted还是非常开心的。

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        map<int,int> eleTimes;
        int temp=0,count=0,sta=0;
        sta=nums.size()/2;
        if(sta==0) return nums[0];
        for(auto i=nums.begin();i!=nums.end();i++){
            temp=(*i);
            ++eleTimes[temp];
        }
        for(auto j=eleTimes.begin();j!=eleTimes.end();j++){
            count=(*j).second;
            if(count>sta) return (*j).first;
        }    
        }
        
    
};

---

后来还看了一种比较简单的思想，但是理解起来比较困难

[C++ Solution] [O(n) computation/O(1) space] The problem can be extended to ⌊n/k⌋ situation

Find k different element, and "remove" them as a group, the remaining element must be the element that appears more than ⌊n/k⌋ times. (Detailed explanation is given in comment)

In this problem, k equals to 2.

Thus we "remove" each pair of 2 different elements, and the remaining element that do not have its counterpart is the desired element. 

就是成k对去“删除”不相同的元素：

例如5个元素的时候，5,3,4,3,3这种情况下，5，3会被删除，4,3会被删除，最后就剩下3；

5,3,3,4,3:5,3会被删除，3,4会被删除，留下3

5,3,3,3,4：5,3会被删除，candidate是第二个3，因为接下来的元素还是3,于是Ntimes变成了2，即使后面的4与其不相同，Ntimes没有被减至0，于是，被返回。

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)
            {
                candidate = num[i];
                nTimes = 1;
            }
            else
            {
                if(candidate == num[i])
                    nTimes ++;
                else
                    nTimes --;
            }
        }
        return candidate;
    }
};