Sort 1-Simple selection sort

1. Choose the basic idea of sorting :
   Scan n Records, select the smallest record to output it, then scan the remaining N-1 records, select the smallest record to output, ... Repeat the process until the last record is left.

1 //1. Simple selection of sorting2  voidSelectsort (intA[],intnum)3  {4      intmin=0, temp=0;//record the minimum number of the corresponding subscript5       for(intI=0; i<num-1; i++)6      {7min=i;8           for(intj=i+1; j<num;j++)9          {Ten              if(a[min]>A[j]) One              { Amin=J; -              } -          } the          if(min!=i) -          { -temp=A[i]; -a[i]=A[min]; +a[min]=temp; -          } +      } A  } at  -  - intMain () - { -     intarray[9]={0}; - Srand (Time (NULL)); in      for(intI=0;i<9; i++) -     { toArray[i]=rand ()% -;//generate 0-99 of random numbers +     } -cout<<"array prior to sorting \ n"; thePrint (Array,9); *Selectsort (Array,9); $cout<<"array after sorting \ n";Panax NotoginsengPrint (Array,9); -  theSystem"Pause"); +     return 0; A}

Operation Result:

2. Calculation of time complexity
The first trip requires a n-i comparison, so the number of comparisons is n (n-1)/2. Exchange times, preferably exchange 0 times, the worst exchange times is n-1 times. The final sort count is the sum of the comparisons and the number of exchanges. As a result, the final total time complexity is still not (N2).

Sort 1-Simple selection sort