"Leetcode" Majority Element (Easy) (*^__^*)

Given an array of size n, find the majority element. The majority element is the element, the appears more than times ⌊ n/2 ⌋ .

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

Ideas:

Find the main element, record the main letter with Major, n record the number of occurrences of major relative to other letters

if n = = 0, set the current number as the primary number

else if the current number equals the primary number, n++

Else the current number is not equal to the primary number, n--

Since the main number appears more than half, the final major must be the main figure.

No tests, just one AC.

classSolution { Public:    intMajorityelement (vector<int> &num) {        intMajor; intn =0;  for(inti =0; I < num.size (); i++)        {            if(n = =0) {Major= Num[i]; n++; }            Else if(Major = =Num[i]) {N++; }            Else{n--; }        }        returnMajor; }};

"Leetcode" Majority Element (Easy) (*^__^*)