sort-algorithm | 史上最全经典排序算法总结(Java实现) | Learning library

After thinking a little more and making some attempts, I realized that the solution is very simple. For each additional criteria, just add a condition block comparing if the properties of the previous block are equal and repeat the same comparison of the previous block for the new properties. In my case, the code looks like this:

When you call merge_sort_ascend recursively, you ignore its return value. That return value is important.

The warning is self explanatory: color.convert_colorspace(in_hsv_h, 'HSV', 'RGB') is of type float64, and imsave, convert elements to uint8.

The pixels of PNG image, are stored as one byte per component (one byte for red, one for green and one for blue).
Each component is an integer value in range [0, 255] (type uint8).

The output of color.convert_colorspace is of float64, each color component is in range [0, 1] of type float64 (stored as 64 bits in memory, and much more accurate than uint8).

The conversion from float64 range [0, 1] to uint8 range [0, 255] is performed like: uint8_val = round(float64_val*255).
The rounding operation loose some data (for example: in case float64_val*255 = 132.658, the result is rounded to 133).

Convert image to uint8 prior to saving to suppress this warning

Tells you to convert the image elements to uint8 prior to saving.

Solution is simple.
Multiply by 255, and add .astype(np.uint8).

Example C code. It uses fewer local variables than the example linked to in the original post, allowing a C compiler to use registers for those variables. This program takes less than 0.5 second to sort 20 million (20*1024*1024 = 20971520) 64 bit unsigned integers by the upper 32 bits on my system (Intel 3770K 3.5ghz cpu, Windows 7 Pro 64 bit).

When you try to access collection[j], the compiler returns a &mut String because that's the type of the vector's elements. When you try to access collection[least_element], the borrow checker doesn't know if least_element != j, and having two mutable references of the same element would be undefined behavior. You can either use std::ops::Index which returns a &&mut String (and it's safe to have two immutable references to the same mutable reference), directly borrowing the elements (&collection[j] < &collection[least_element]) or, if possible, changing the type of collection to Vec<&String> or Vec.

This should work. It also uses range() to avoid having to use a while loop.

The array array is initialized when an object of the type InputReader is instantiated, not after the readInput() method is run.

In other words, array is always initialized when you use new InputReader(), not after the readInput() method is run. That means that the arrayLength variable has not yet been modified at the moment when array is created, so the default value for an int is used, which is 0.

Try this:

There must be some issue with your code itself. please share the code for better investigation.

Also If you want to check out the working natural sort algorithm implementation.

excerpt:

Yes, there is. Basically, we can make the inequalities stricter: change a[l]<=p to a[l]

and a[r]>=p to a[r]>p. This will help handle the case when there can be equal elements in the array, too, so you can get rid of the l and left conditions.

The pivot is not guaranteed to land at precisely position l, then, but still, a[left...l] will be <=p and a[l+1...right] will be >=p, which is just enough for the algorithm to work.

Here is a variation of your code that works. The condition if(l < r) is modified to be if(l <= r), and inside, l++; and r--; are added to skip the two elements we just swapped. Also, the recursive calls are updated to (left,r) and (l,right), since there is now no guaranteed position for the pivot element.