Fast row (c ++)

Worst case time complexity of n- ² , the average nlogn

Explanation: traversing a O (n), fast row using divide and conquer to traverse, as a binary tree, the number of times it traversed a relationship with his depth, n nodes, then the depth is at least log (n + 1), so the average nlogn; N binary tree that is the maximum depth, so the worst case time complexity of n- ² .

// Get a reference number, and into the middle
 // this function is performed once, a reference number to find the correct position and to put 
int mid_index (ARR [], left, right) {
     int I = left + . 1 ;
     int J = right;
     int TEMP = ARR [I];
     the while (I <= J) {
         the while (ARR [I] <TEMP) I ++ ;
         the while (ARR [J]> TEMP) J, ;
         IF (I < J) the swap (ARR [I], ARR [J]); 
    } 
    // to the borders of 
    the swap (ARR [J], ARR [left]);
     return J; 
} 


void quick_sort (ARR [], left, right) {
     // ensure length of at least 2
    if(left>=right) return;
    int mid=mid_index(arr,left,right);
    quick_sort(arr,left,mid-1);
    quick_sort(arr,mid+1,right);
}

Boundary Analysis (first while loop condition i <= j):

When i = j, the right arr center, this time arr [i] == arr [j]:

• If the value is less than the temp, i ++ i to the first number is greater than temp, j of the same exchange was arr [j], and values ​​arr [left] j are to the left than the reference value, the reference value is greater than the right , that the reference value to the correct location of the
• If arr [i] == arr [j]> temp, i is not judged, J, after the last number than the reference value, when i> j, exchange arr [j] and arr [left], supra

 

 

 

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