import java.util.Scanner;
public class PairsOfElementsSumEqualsN {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner s = new Scanner(System.in);
System.out.println("enter the number");
int n = s.nextInt();
int[] array = new int[]{1,2,1,2,4,5,6,7,8,9};
int j = 0;
int i = 0;
for(i = 0 ; i<array.length;i++)
{
for(j=i+1;j<array.length;j++)
{
if(array[i] + array[j] == n)
{
System.out.println(array[i] + "," + array[j]);
}
}
}
}
}
i think it should be n^2
but i want an explanation for the answer
4 个答案:
答案 0 :(得分：5)
Yes you are correct its time complexity is O(n^2).
You are doing n-1 comparisons in 1st pass, n-2 in 2nd pass, n-3 in 3rd pass and so on. So the total number of comparisons will be.
(n-1)+(n-2)+(n-3)+.....+3+2+1
Sum = n(n-1)/2
i.e O(n^2)
This is because big-O notation describes the nature of the algorithm. The major term in the expansion (n-1) * (n-2) / 2 is n^2. And so as n increases all other terms become insignificant.
答案 1 :(得分：1)
yes it is n^2 because your both loops iterate over the entire limit and hence you have n times first loop into (n-1) times second loop leads to n^2.
答案 2 :(得分：0)
Just count the operations
Outer loop 1st
Inner Loop N-1 times
print and if body are constant i.e O(1)
Outer loop 2nd
Inner Loop N-2 times
print and if body are constant i.e O(1)
...
Outer loop N-1th time
Inner Loop 1 times
print and if body are constant i.e O(1)
Outer loop N-1 time
Inner Loop 0 times
print and if body are constant i.e O(1)
Total (N-1)+(N-2)+...+1
=N^2-N(N+1)/2
答案 3 :(得分：0)
basically as @SSH said it's the worst case scenario, even if you have a constant case where, eg, you do that if block for even Js, the result of the algorithm would be nn/2 but for n going to infinite that "divide by 2" is irrelevant.
For the same reason every addition to the complexity (nn/2 + 5 or n*n -5) is irrelevant as n is gonna be "pushing" to infinite