1复杂度分析

1 int cal(int n) {
2    int sum = 0;
3    int i = 1;
4    for (; i <= n; ++i) {
5      sum = sum + i;
6    }
7    return sum;
8 }

 1 int cal(int n) {
 2    int sum = 0;
 3    int i = 1;
 4    int j = 1;
 5    for (; i <= n; ++i) {
 6      j = 1;
 7      for (; j <= n; ++j) {
 8        sum = sum +  i * j;
 9      }
10    }
11 }
12  

1 i=1;
2 while (i <= n)  {
3    i = i * 2;
4 }

 1 void print(int n) {
 2   int i = 0;
 3   int[] a = new int[n];
 4   for (i; i <n; ++i) {
 5     a[i] = i * i;
 6   }
 7  
 8   for (i = n-1; i >= 0; --i) {
 9     print out a[i]
10   }
11 }

 1 void print(int n) {
 2   int i = 0;
 3   int[] a = new int[n];
 4   for (i; i <n; ++i) {
 5     a[i] = i * i;
 6   }
 7  
 8   for (i = n-1; i >= 0; --i) {
 9     print out a[i]
10   }
11 }