Time complexity with conditional statements

问题

How does one calculate the time complexity with conditional statements that may or may not lead to higher oder results?

For example:

for(int i = 0; i < n; i++){  
   //an elementary operation   
   for(int j = 0; j < n; j++){
       //another elementary operation  
       if (i == j){  
           for(int k = 0; k < n; k++){
               //yet another elementary operation
           }
       } else {
           //elementary operation
       }
   }
}

And what if the contents in the if-else condition were reversed?

回答1:

Your code takes O(n^2). First two loops take O(n^2) operations. The "k" loop takes O(n) operations and gets called n times. It gives O(n^2). The total complexity of your code will be O(n^2) + O(n^2) = O(n^2).

Another try:

 - First 'i' loop runs n times.
 - Second 'j' loop runs n times. For each of is and js (there are n^2 combinations):
     - if i == j make n combinations. There are n possibilities that i==j, 
      so this part of code runs O(n^2).
     - if it's not, it makes elementary operation. There are n^2 - n combinations like that
       so it will take O(n^2) time.
 - The above proves, that this code will take O(n) operations.

回答2:

That depends on the kind of analysis you are performing. If you are analysing worst-case complexity, then take the worst complexity of both branches. If you're analysing average-case complexity, you need to calculate the probability of entering one branch or another and multiply each complexity by the probability of taking that path.

If you change the branches, just switch the probability coefficients.

来源：https://stackoverflow.com/questions/37965609/time-complexity-with-conditional-statements