算法的复杂度和Big-O

Question

So I am preparing for an exam and 25% of that exam is over Big-O and I'm kind of lost at how to get the complexity and Big-O from an algorithm. Below are examples with the answers, I just need an explanation of how to the answers came to be and reasoning as to why some things are done, this is the best explanation I can give because, as mentioned above, I don't know this very well:

    int i =n;  //this is 1 because it is an assignment (=)
    while (i>0){                //this is log10(10)*(1 or 2) because while 
        i/=10;  //2 bc / and = // loops are log base (whatever is being /='d
    } //the answer to this one is 1+log10(n)*(1 or 2) or O(logn)
      //so i know how to do this one, but im confused when while and for 
      //loops nested in each other

    int i = n; int s = 0;
    while (i>0){
        for(j=1;j<=i;j++)s++;{
        i/=2;
    } //the answer to this one is 2n +log2(n) + 2 or O(n)
      //also the i/=2 is outside for loop for this and the next one

    int i = n; int s=0
    while (i>0){
        for(j=1;j<=n;++J) s++;
        i/=2;
    } //answer 1+nlogn or O(nlogn)

    int i = n;
        for(j=1;j<=n;j++)
        while(i>o) i/=2; 
           //answer is 1+log2(n) or O(log(n))


    for(j=1; <=n; ++j){
        int i-n;
        while(i>0) i/=2;
    } //answer O(nlog(n))

Answer 1

Number 4: the for loop counts from 1 to N, so it is at least O(n). The while loop takes O(log n) the first time, but since i doesn't get reset, while loop has only has one iteration each successive time through the for loop. So basically O(n + log n), which simplifies to O(n).

Number 5: same as above, but now i does get reset each time, so you have O(log n) done N times: O(n log n).