我只是不明白，为什么时间复杂度是O(n ^ 2)而不是O(n * logn)? 第二个循环每次递增2，所以它不是O(logn)吗?
I just didn't get it, why the time complexity is O(n^2) instead of O(n*logn)? The second loop is incrementing 2 each time so isn't it O(logn)?
void f3(int n){ int i,j,s=100; int* ar = (int*)malloc(s*sizeof(int)); for(i=0; i<n; i++){ s=0; for(j=0; j<n; j+=2){ s+=j; printf("%d\n", s); } free(ar); }推荐答案
通过递增2而不是1，您正在执行以下N*N*(1/2).使用big(O)表示法时，您无需关心常数，因此它仍然是N * N.这是因为big(O)表示的是算法增长的复杂性.
By incrementing by two, rather than one, you're doing the following N*N*(1/2). With big(O) notation, you don't care about the constant, so it's still N*N. This is because big(O) notation reference the complexity of the growth of an algorithm.