[day_030420] numbers that appear more than half of the time in the array

A number that appears more than half the times in an array

Title source

Cattle from ： A number that appears more than half the times in an array

Title Description

Give a length of n Array of , There is a number in an array that appears more than half the length of the array , Please find out the number .
For example, enter a length of 9 Array of [1,2,3,2,2,2,5,4,2]. Because of the number 2 Appears in the array 5 Time , More than half the length of the array , So output 2.

Data range ：n≤50000, The value of the element in the array 0≤val≤10000
requirement ： Spatial complexity ：O(1), Time complexity O(n)

Input description

The first line of input is a positive integer n(1 ≤ n ≤ 10^5)
The second line includes 3*n It's an integer a_i(1 ≤ a_i ≤ 10^9), Represents the level value of each contestant .

Example 1

Input

[1,2,3,2,2,2,5,4,2]

Return value

2

Example 2

Input

[3,3,3,3,2,2,2]

Return value

3

Example 3

Input

[1]

Return value

1

Thought analysis

Solution 1

• Offset method
• If two numbers are not equal , Just eliminate these two numbers , In the worst case , Eliminate one mode and one non mode at a time , So if there are modes , The last number left must be the mode
• Define a ret Save the current mode , Definition times Indicates the number of occurrences
• Traversal array , The current array value is the same as ret Equal increases times value , Inequality decreases times value ,times The value is 0 Replace with the value of the current array element ret

Solution 2

• Sort the array
• Find the number in the middle of the array , Because if a number exists in the array more than half times , Then the number in the middle of the array after sorting must be this number

Code display

Solution 1

class Solution {
    
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) {
    
        int times = 1;
        int ret = numbers[0];// Definition ret For the first element of the array , If an array has only one element , Can return directly ret
        int sz = numbers.size();
        for (int i = 1; i < sz; i++)
        {
    
            // At present times Value is not 0
            if (times != 0)
            {
    
                if (numbers[i] != ret)
                {
    
                    times--;
                }
                else
                {
    
                    times++;
                }
            }

            // At present times The value is 0, The original ret Has been offset by unequal values 
            else
            {
    
                // Assign the value of the current array to ret,tims Reset to 1
                ret = numbers[i];
                times = 1;
            }
        }
        return ret;
    }
};

Solution 2

class Solution {
    
public:
    int MoreThanHalfNum_Solution(vector<int> numbers) {
    
        sort(numbers.begin(),numbers.end());
        int sz=numbers.size();
        int max=numbers[sz/2];// The middle of the array 
        return max;
    }
};