How to use C++ for algorithm Analysis
In this issue, the editor will bring you about how to use C++ for algorithm analysis. The article is rich in content and analyzes and narrates it from a professional point of view. I hope you can get something after reading this article.
Time complexity
O (1) < O (logn) < O (n) < O (n ^ 2) < O (n ^ C) < O (C ^ n)
Constant < logarithmic order < linear order < square order < polynomial order < exponential order
Constant order
Void aFunction () {
Int c = 10 + 20
Int d = c * c; printf (d)
}
Analysis: the amount of calculation is 2. 2 is a constant function, so it has little effect on the growth of constant function, so it is recorded as O (1).
Linear order
Void bFunction (int n) {
For (int I = 0 ~ (th) I < n ~ (th) I ~ +) {/ / n
Int c = 2 * I; / / 1
Int d = 3 * I; / / 2
}
}
Analysis: the amount of calculation of the function is equal to (n) (2), and the 2 constant is negligible, so it is recorded as O (n).
Square order
Void bFunction (int n) {
For (int I = 0 ~ (th) I < n ~ (th) I ~ +) {
For (int j = 0 * * j < I * * j +) {
}
}
}
Analysis: the amount of calculation of the whole function is (n) (n ^ 1), and the constant quantity is ignored, so it is written as O (n ^ 2).
Logarithmic order
Void bFunction (int n) {
For (int I = 3 * * I < n;) {
I * = 3;}
}
Analysis: suppose that the condition of the cycle s times is s = 3 ^ s < n; the logarithm is expressed as s = log3n, marked as O (log3n), and the constant can be ignored O (logn).
Multinomial order
Void bFunction (int n) {
For (int I = 0 ~ (th) I < n ~ (th) I ~ +) {
For (int j = 0 * * j < n * j +) {
For (int k = 0 * * k < n * k +) {
}
}
}
}
Analysis: the amount of calculation is n ^ 3, and the power is constant and recorded as O (n ^ C).
Exponential order
Void bFunction (int n) {
Int num = n
For (int I = 0 ~ (th) I < n ~ (th) I ~ +) {/ / O (n)
Num * = n
}
For (int j = 0 position j)
