Write a function, statistics of the given points in the array, is higher than each score, the number of the number of array by a new one dimensional array (dynamic application) need to return, because of the array and the original score array length, so don't need to return the length of the new array, you can control the time complexity O (N)? Note: you can think the score is zero to 100,

Input the sample 1:
The first line is the number of the second line with as some of the scores, space-delimited
6
80, 70, 60, 90, 79, 80
The output sample 1:
Calculates and returns the number of points higher than the corresponding position, the junction should be returned as new array, the output by the test procedure is complete,
,4,5,0,3,1 {1}
Higher than 80 points of 1:90, more than 70 points has four: 80,90,79,80... Is implicit illustrates its score ranking position... ,

Enter the sample 2:
5
80 80 80 90 88
The output sample 2:
Calculates and returns the number of points higher than the corresponding position, is the address at the beginning of a new array (dynamic application space), the output by the test procedure is complete,
,2,2,0,1 {2}
Function interface definition:
Int * fun (int * d, int N);
/* please fill in the answer here */

CodePudding user response:

Knock this is my code, but not time complexity, there is a test point time, how to improve??
How to use a loop to complete??

 int * fun (int * d, int N) {
int i,j; 
Int * a=(int *) malloc (sizeof (int) * N); 
for(i=0; ifor(j=0; jIf (d [j] & gt; D [I]) 
A [I] + +; 
} 
return a; 
} 

CodePudding user response:

 int m=6; 
Int arr [] {80, 70, 60, 90, 79,}; 
Int result [101]; 
Memset (result, 0, sizeof (int) * 101); 
for (int i=0; I & lt; 6; I++) 
{
for (int j=0; J & lt; Arr [I]; J++) 
The result [j] + +; 
} 
For (int m=0; M & lt; 6; M++) 
Printf (" % d ", the result [arr [m]]).