for(int i=0;i<n;i++)
{
// Some code
}
We generally say that this loop runs for n+1 times so n+1 steps for this and there is one step is of initialisation i=0. This I have read in most of the text books. My question is this that everytime the loop runs there is one more step of incrementing i to i+1 that is i=i+1, this is also one of the step which should be counted in calculating the time complexity.As I am a newbie to algorithm analysis help me with this problem.
我们一般说来,这一轮机为零+1倍,为这一[......]步骤。
无,我们说的是n。 这在将<代码>0的起始指数与作为<代码>和t;n书写的边界-核对数相结合方面起了重要作用。 反射线一旦到达n <>t>后,将退出该航程(0,1,1,2,......1,0,1,0,1,0,1,0,0)。
为加固该柜台而做的工作,即:i++>/code>、i += 1或i = compute_the_next_index(i),并不算作“步骤”。 这些步骤是劝说,即处决尸体。
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