// returns true if the line from (a,b)->(c,d) intersects with (p,q)->(r,s) function intersects(a,b,c,d,p,q,r,s) { var det, gamma, lambda; det = (c – a) * (s – q) – (r – p) * (d – b); if (det === 0) { return false; } else { lambda = ((s – q) * (r – a) + … Read more
Arash Partow’s FastGEO library contains implementations of many interesting algorithms in computational geometry. Polygon intersection is one of them. It’s written in Pascal, but it’s only implementing math so it’s pretty readable. Note that you will certainly need to preprocess your edges a little, to get them into clockwise or counterclockwise order. ETA: But really, … Read more
Set.retainAll() is how you find the intersection of two sets. If you use HashSet, then converting your ArrayLists to Sets and using retainAll() in a loop over all of them is actually O(n).
Using HashSet<Character>: HashSet<Character> h1 = new HashSet<Character>(), h2 = new HashSet<Character>(); for(int i = 0; i < s1.length(); i++) { h1.add(s1.charAt(i)); } for(int i = 0; i < s2.length(); i++) { h2.add(s2.charAt(i)); } h1.retainAll(h2); Character[] res = h1.toArray(new Character[0]); This is O(m + n), which is asymptotically optimal.
Place both series in Python’s set container then use the set intersection method: s1.intersection(s2) and then transform back to list if needed. Just noticed pandas in the tag. Can translate back to that: pd.Series(list(set(s1).intersection(set(s2)))) From comments I have changed this to a more Pythonic expression, which is shorter and easier to read: Series(list(set(s1) & set(s2))) … Read more
The algorithm is as follows: the smaller set is looped over and every element is copied depending whether it’s found in the bigger set. So, it’s the C equivalent of def intersect(a, b): if len(a) > len(b): a, b = b, a c = set() for x in a: if x in b: c.add(x) return … Read more
It’s possible to check if a point is inside a sector with only integer arithmetic and the basic operations of addition, subtraction and multiplication. For a point to be inside a circular sector, it has to meet the following tests: It has to be positioned counter-clockwise from the start “arm” of the sector. It has … Read more
Most of the answers already here seem to follow the general idea that: find the intersection of two straight lines passing the given points. determine if the intersection belong to both line segments. But when intersection does not occur often, a better way probably is to reverse these steps: express the straight lines in the … Read more
Yes there are differences which may or may not be relevant in your scenario. Perhaps the most significant is the difference in how members with the same property key are handled when present in both types. Consider: interface NumberToStringConverter { convert: (value: number) => string; } interface BidirectionalStringNumberConverter extends NumberToStringConverter { convert: (value: string) => … Read more
Use NSMutableSet: NSMutableSet *intersection = [NSMutableSet setWithArray:array1]; [intersection intersectSet:[NSSet setWithArray:array2]]; [intersection intersectSet:[NSSet setWithArray:array3]]; NSArray *array4 = [intersection allObjects]; The only issue with this is that you lose ordering of elements, but I think (in this case) that that’s OK. As has been pointed out in the comments (thanks, Q80!), iOS 5 and OS X 10.7 … Read more