数据结构----算法复杂度分析

int i = 0;int n = 100; 
printf("test"); 
printf("test");
printf("test"); 
printf("test"); //算法时间复杂度为O(1)

int i = 0;int n = 100; int sum = 0;
for(i=0; i<n; i++)
{
    sum = sum + i;
}
//算法时间复杂度为O(n)

int i = 0, j = 0;int n = 100; int sum = 0;
for(i=0; i<n; i++)
{
    for(j=0; j<n; j++)
    {
        printf("ok");
    }
}
//算法时间复杂度为O(n^2)

for(i=0; i<n; i++)
{
    for(j=i; j<n; j++)
    {
        printf("ok");
    }
}//n + n-1 + n-2 + ... + 1 = n*(n+1)/2 ---> O(n^2)
//算法时间复杂度为O(n^2)

int i = 1; int n = 100;
while(i < n)
{
    i = i * 2;
}//2^x = n --->x = log(n)
//算法时间复杂度为O(log(n))