"Leetcode" Majority Element

Test instructions

Given an array of size n, find the majority element. The majority element is the element, which appears more than? N/2? Times.

Assume that the array was non-empty and the majority element always exist in the array.

Ideas:

Find the number of occurrences greater than N/2 in the array.

In C + +, map is particularly handy for this number of statistics, and Python is of course used in dictionaries.

Code:

C++

Class Solution{public:    int majorityelement (vector<int> &num) {        Vector<int>::iterator it = Num.begin ();        int n = num.size ();        map<int,int> count;        for (; it! = Num.end (); it++)        {            //Why is there no need to determine the existence of a key here? Initialize to 0            if (++count[*it] > N/2)            {                return *it;            }}        } because    it does not exist;

Python:

Class solution:    # @param num, a list of integers    # @return An integer    def majorityelement (self, num):        coun t = Dict ([])        flag = len (num)/2 for        item in num:            if item in count:                count[item] = Count[item] + 1            else:< C9/>count[item] = 1                            if Count[item] > flag:                return item

"Leetcode" Majority Element