什么是执行时间增长率这个代码的大O?

时间:2012-04-26 22:22:49

标签: time-complexity big-o

这段代码的执行时间增长率大O是什么?

int maxValue = m[0][0];         
for (int i = 0; i < N; i++)         
{                       
    for (int j = 0; j < N; j++)         
    {                       
        if ( m[i][j] > maxValue )           
        {                       
            maxValue = m[i][j];         
        }                       
    }                       
}                   
cout << maxValue << endl;           
int sum = 0;                    
for (int i = 0; i < N; i++)         
{                       
    for (int j = 0; j < N; j++)         
    {                       
        sum = sum + m[i][j];            
    }                       
}                           
cout << sum << endl;                            

1 个答案:

答案 0 :(得分:0)

O(N ^ 2),因为你有两个嵌套的“for”循环,从0到N.