kandi X-RAY | swift-algorithm-club Summary
kandi X-RAY | swift-algorithm-club Summary
Here you'll find implementations of popular algorithms and data structures in everyone's favorite new language Swift, with detailed explanations of how they work.
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Trending Discussions on swift-algorithm-club
I've been studying up on graphs using the Adjacency List implementation, and I am reading that adding an edge is an O(1) operation.
This makes sense if you are just tacking an edge on to the Vertex's Linked List of edges, but I don't understand how this could be the case so long as you care about removing the old edge if one already exists. Finding that edge would take O(V) time.
If you don't do this, and you add an edge that already exists, you would have duplicate entries for that edge, which means they could have different weights, etc.
Can anyone explain what I seem to be missing? Thanks!...
ANSWERAnswered 2018-Aug-21 at 06:22
You're right at your complecxity analysis. Find if edge already exist is truly O(V). But notice that adding this edge even if existed is still O(1).
You need to remember that having 2 edges with the same source an destination are valid input to graph - even with different weights (maybe not even but because).
That way adding edge to adjacency-list-graph is O(1)
Based on this Ray Wenderlich article I'm able to create a binary tree data structure like that:
ANSWERAnswered 2019-Apr-08 at 00:04
In terms of the basics of how to draw a line, you:
- Move to the starting point with
- Adding line to the end point with
You can then render that path in your UI by either:
CAShapeLayer, specifying its
fillColor; set its
path, and then add that shape layer as a sublayer of the view’s
UIViewsubclass and in its
draw(_:)method you can call
setStrokeof the desired
UIColor, set the
UIBezierPath, and then
Generally, I’d use the
CAShapeLayer approach, where I basically configure the shape layer, but let the OS render that shape layer for me.
That having been said, I’d probably take this a step further, and wrap the line drawing in its own
UIView subclass. The thought process is that not only are high-level views generally composed of
UIView objects, but it also opens the door for all sorts of advanced UX (e.g. you might want to detect a tap on a node and do something).
Anyway, I’d wrap that “connector line” drawing code in a
UIView subclass, like so:
Going through Swift algorithms including the brute force string search i.e.: https://github.com/raywenderlich/swift-algorithm-club/tree/master/Brute-Force%20String%20Search
The output is said to be 7
Now actually the output is (String.Index?) - and although I can subscript with that I want to actually see the value - 7.
The code is...
ANSWERAnswered 2019-Feb-26 at 03:06
Try this out:
ANSWERAnswered 2018-Apr-17 at 18:46
My solution is a little bit more wordy :) where I replaced printTree with my own two methods
I have code like this:...
ANSWERAnswered 2018-Mar-11 at 04:26
Map your generic type
String and then apply
joined is only for elements of type
String and here compiler do not know the type of
I have a series of coordinates that I'm wanting to find the linear regression for, so I can then find the bearing of the line. I'm using the Linear Regression algorithm from Swift Algorithm Club on this set of coordinates:...
ANSWERAnswered 2017-Sep-17 at 08:55
I confirm that your bearing function works correctly. However, the regression value you get for latitude is 51.45437262420372 - which is where the regression line intersects with longitude = 0.0. I think you need to use both the intersection point and slope of the regression line to calculate fitted lat,lon values (within your map region), and then find the bearing using those values.
LATER ADDITION: An important issue here is that longitude coordinates are not linear in relation to distance - the distance covered by a degree of longitude varies with latitude. I think the simplest solution is just to convert your array of lat,lon values to lat,normalisedLon values where normalisedLon = lon * cosine(lat).
When you do this the angle of your path from your first coordinate to your last coordinate = 3.055 degrees (just east of north which appears to be correct).
Once you've done this then the slope of your regression line will in fact represent the direction you are seeking i.e. heading = atan(slope).
However, your regression code just doesn't seem to work for me. It should give a result very close to 3 degrees also, but it is way off.
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