这个for循环的大O分析

Question

 sum = 0;
 for (i = 1; i <= n; i++) {    //#1
   for (j = 1; j <= i * i; j++) {     //#2
      if (j % i == 0) {    //#3 
          for (k = 1; k <= j; k++) {   //#4
             sum++;
         }
     }
  } 

}

以上让我感到困惑

Suppose #1 runs for N times
    #2 runs for N^2 times
    #3 runs for  N/c since for N inputs N/c could be true conditions
    #4 runs for  N times

因此大致我可以看O(N ^ 5).我不确定.请帮忙澄清一下.

编辑我想知道运行时if(j%i==0).由于它N^2从其父循环中获取输入,因此可以(N^2)/c执行而不是执行N/c

Answer 1

我会说它的O(N ^ 4)和它一样.

 for (int i = 1; i <= n; i++)        //#1 O(n ...
   for (int j = i; j <= i * i; j+=i) //#2 ... * n ...
     for (int k = 1; k <= j; k++)    //#4 ... * n^2) as j ~= i^2
         sum++;

要么

public static void main(String... args) {
    int n = 9000;
    System.out.println((double) f(n * 10) / f(n));
}

private static long f(long n) {
    long sum = 0;
    for (long i = 1; i <= n; i++)   //#1
        for (long j = 1; j <= i; j++) //#2
            sum += i * j; // # 4
    return sum;
}

版画

9996.667534360826

这非常接近10 ^ 4