Data Structures and Algorithms based computation time complexity ------

Efficiency of the algorithm from the following two aspects:

1. Time Efficiency: it refers to the time it takes for the algorithm
2. Space Efficiency: refers to the process spent executing the algorithm in the storage space

Time efficiency and space efficiency are sometimes contradictory.
Here we only discuss pre-analysis, post hoc analysis method also because the objective conditions and other computer hardware and software and so on.
Prior analysis
run time of the algorithm is substantially equal to a computer to perform a simple operation (such as assignment, comparing, moving, etc.) and the time required to perform the algorithm is simple product number .
Matrix, for example:

for(i=1;i<=n;i++)		//n+1次
	for(j=1;j<=n;j++)	//n*(n+1)次
	{
		c[i][j]=0;		//n*n次
		for(k=0;k<=n;k++)	//n*n*(n+1)次
			c[i][j]=c[i][j]+a[i][k]*b[k][j];//n*n*n次
	}

Then the time consumed by the above algorithm is that the algorithm execution times each statement sum, the time consumed T (n-) = 2N ^{. 3} + with 3N ² + 2N +. 1.
In order to facilitate comparison of different time efficiency of the algorithm, we only compare their magnitude.

• If an auxiliary function f (n) (i.e., containing only a function of the highest magnitude), such that when n approaches infinity, T (n) / f ( n) is the limit value not equal to zero constants , called f ( n) is T (n) is a function of the same order. Denoted by T (n-) = O (f (n)) , said O (f (n)) is the time complexity.

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Analysis algorithm time complexity:

• Find out the statement is the maximum frequency that statement (that is, the maximum number of execution algorithm) as the basic statement
• Calculation of basic statement of the frequency of the resulting problem size n of a function f (n)
• Whichever is highest magnitude represented by symbol O

Time complexity is determined by the frequency of nested deepest statement.
example:

void exam(float x[][],int m,int n)
{
	float sum[];
	for(int i=0;i<m;i++)
	{
		sum[i]=0.0;
		for(int j=0;j<n;j++)
			sum[i]+=x[i][j];//嵌套最深层
	}

example:

i=1;	
while(i<=n)
	i=i*2;

Set the number of executions of X
2 ^X <= n-
X <= log ₂ n-
execute the maximum number of log ₂ n-, T (n-) = O (log ₂ n-)