C++ 时间复杂度

看到网上一些资料的案例不全，所以自己开个来复习。

O(1)＜O(log2n)＜O(n)＜O(nlog2n)＜O(n^2)＜O(n^3)＜…＜O(2^n)＜O(n!)

 

O(1)

int a = 1, b = 3, c = 3;

O(log2n)

int i = 1;
while(i<=n)
{
    i*=2
}

O(n)

int sum = 0;
for(int i = 0; i < n; i++)
{
    sum+=i;
}

O(nlog2n)

for(int m=1;m<n;m++)
{
    i=1;
    while(i<n)
    {
        i=i*2;
    } 
}

O(n^2)

int num1, num2;
for(int i=0; i<n; i++)
{ 
     num1 += 1;
     for(int j=1; j<=n; j++)
    { 
         num2 += j;
     }
} 

O(n^3)

int num1, num2, num3;
for(int i=0; i<n; i++)
{ 
     num1 += 1;
     for(int j=1; j<=n; j++)
     { 
        num2 += j;
        for(int k=1; k<=n; k++)
        {
             num3 += k;
        }         
     }
} 

O(2^n)

2^n = 2*2*2*........2* // 有n个2

O(n!)

n! = 1*2*3*4*5......*n;

相关文章: https://blog.csdn.net/u012925946/article/details/84030160

                https://blog.csdn.net/qq_41672557/article/details/101699850

                https://blog.csdn.net/qq_40513633/article/details/107705801