第二章 Big O notation 试题以及讲解 （包会）

这里列出了一些难做的题或是易错的题，简单的我就没有过多的解释，但是如果大家有任何问题都可以私信我或是评论一下，我会尽量即时的解答问题或是疑问的。

int x=0; 

 for(int i=4*n; i>=1; i--) 

     x=x+2*i;

   O(n)

The loop runs O(n) times and does O(1) work per iteration.

int z=0;

int x=0;

for (int i=1; i<=n; i=i*3){ 

   z = z+5;

   z++;

   x = 2*x;

}

O(log n)

Think about the values of i as the loop progresses. It will take on the series of values 1, 3, 9, 27, 81, 243, ..., 3k. Since i is tripling on each iteration, it takes on successive powers of three.

The loop clearly only does O(1) work per iteration, so the main 

question here is how many total iterations there will be. The loop 

will stop when i > n. If we let k be some arbitrary iteration of the 

loop, the value of i on iteration k will be 3k. The loop stops when 

3k > n, which happens when k > log3 n. Therefore, the number of 

iterations is only O(log n)

int y=0;

 for(int j=1; j*j<=n; j++) 

 j <= (n)^1/2 y++;

O(√n)

Notice that j is still growing linearly, but the loop runs as long as 

j2 ≤ n. This means that as soon as j exceeds √ n, the loop will 

stop. Therefore, there will only be O(√n) iterations of the loop, 

and since each one does O(1) work, the total work done is O(√n)

int b=0; //constant

for(int i=n; i>0; i--) 

   for(int j=0; j<i; j++) 

      b=b+5;

O(n^2)

The most accurate answer would be O(n2)

int y=1;

int j=0;

for(j=1; j<=2n; j=j+2) 

   y=y+i;

int s=0;

for(i=1; i<=j; i++) 

   s++;

O(n)

int b=0;

for(int i=0; i<n; i++) 

    for(int j=0; j<i*n; j++)

       b=b+5;

The inner loop will run 0 + n + 2n + 3n + 4n + ... + n(n-1) = n(0 + 1 

+ 2 + ... + n - 1) times, so the total work done is O(n3). You 

shouldn't multiply by the number of times the outer loop runs because 

you're already summing up across all iterations. The most accurate 

runtime would be O(n3)

int t=0;

for(int i=1; i<=n; i++) 

   for(int j=0; j*j<4*n; j++) 

      for(int k=1; k*k<=9*n; k++) 

         t++;

Look at the second loop. This actually runs O(√n) times using the 

same logic as one of the earlier parts. That third inner loop also 

runs O(√n) times, and so the total work done will be O(n2)

int a = 0;

int k = n*n;

while(k > 1) 

{

    for (int j=0; j<n*n; j++) 

       { a++; }

     k = k/2;

}

The outer loop starts with k initialized to n2, but notice that k is 

halved on each iteration. This means that the number of iterations of 

the outer loop will be log (n2) = 2 log n = O(log n), so the outer 

loop runs only O(log n) times. That inner loop does do O(n2) work, so 

the total runtime is O(n^2 log n)

int i=0, j=0, y=0, s=0;

for(j=0; j<n+1; j++) 

   y=y+j;

for(i=1; i<=y; i++) 

   s++;

O(n^3)