问题
My friend has a small problem and I'm at the end of my knowledge. He wrote a Simple (he got it in school) QuickSort Algorithm And it produces a StackOverflow Error. I know it means it calls itself recursive too many times somewhere, but I can't get the logical Error - please help me!
Here is the code (I'm leaving out some code as that is only to show it in 2 text areas):
int array [] = new int [10];
...
public static void quicksort (int array[],int l,int r){
int i = l;
int j = r;
int mitte = array [(l+r)/2];
while (i<j) {
while (array[i]<mitte) {
i++;
} // end of if
while (mitte<array[i]) {
j--;
} // end of if
if (i<=j) {
int merke =array[i];
array[i] = array [j];
array [j] = merke ;
i++;
j--;
} // end of if
if (i<j) {
quicksort(array,l,j);
} // end of if
if (l<r) {
quicksort(array,l,r);
} // end of if
} // end of while
}
It is called like this:
quicksort(array,0,9);
But, if We call it and the 2 numbers are the same, it gives no Overflow.
If more code is needed, here is the full Code on pastebin: http://pastebin.com/cwvbwXEu
回答1:
First, you have an infinite loop here:
while (mitte<array[i]) {
j--;
} // end of if
It needs to be array[j]
Secondly (and leading to the infinite recursion), in the second call to quicksort
if (l<r) {
quicksort(array,l,r);
} // end of if
In recursion, you always need to shorten the range that you call yourself with, or else it'll be infinite. I haven't worked out exactly what you're doing, but I think you mean:
if (i<r) {
quicksort(array,i,r);
} // end of if
回答2:
This call:
if (l<r) {
quicksort(array,l,r);
}
recursively calls quicksort
with the same arguments that were passed in, rather than calling with a smaller subproblem to solve. It will therefore infinitely recurse.
回答3:
if (l<r)
quicksort(array,l,r);
Isnt l always less than r? this will cause an infinite recursion which is why you dont get an overflow if both values are the same.
来源:https://stackoverflow.com/questions/21388639/simple-quicksort-algorithm-giving-stack-overflow-error