Multiple intersection s1&s2&..&sn(n-1)*O(l) where l is max(len(s1),..,len(sn))Differen...
add():添加元素clear():清空集合copy():拷贝一个集合difference:返回多个集合的差集difference_update():移除集合中的元素,该元素在指定的集合也存在discard(obj):删除集合中指定的元素intersection(set1, set2 … etc):返回两个或更多集合中都包含的元素,即交集,返回一个新集合intersection_update():返回两个或更...
(3)集合:set 参考dict,故意实现很相似。 As seen in thesource codethe complexities for set difference s-t or s.difference(t) (set_difference()) and in-place set difference s.difference_update(t) (set_difference_update_internal()) are different! The first one is O(len(s)) (for every e...
(3)集合:set 参考dict,故意实现很相似。 As seen in thesource codethe complexities for set difference s-t or s.difference(t) (set_difference()) and in-place set difference s.difference_update(t) (set_difference_update_internal()) are different! The first one is O(len(s)) (for every e...
set内置的各种函数: add():添加元素 clear():清空集合 copy():拷贝一个集合 difference:返回多个集合的差集 difference_update():移除集合中的元素,该元素在指定的集合也存在 discard(obj):删除集合中指定的元素 intersection(set1, set2 … etc):返回两个或更多集合中都包含的元素,即交集,返回一个新集合 ...
s=set(['a','b']) 特点: 无重复 操作: 查询: ains (返回a是否存在于s集合) s|t s.union(t)(返回s和t并集) s&t s.intersection(t)(返回s和t交集) s-t s.difference(t)(返回s中有t没有的元素集合) 插入: s.add(a)(插入元素a)
).intersection(range(500, 1500))"1000 loops, best of 3: 120 usec per loopList 大概用了Set的...
It’s the part of the green line passing through the gray area from the intersection point with the blue line to the intersection point with the red line. The latter point is the solution. If you insert the demand that all values of x must be integers, then you’ll get a mixed-...
The set intersection operation is generally defined as the number of elements that are common between two sets, represented as sorted arrays of integers. The most common way to compute it is a linear scan: size_t intersection_size(int *a, int *b, size_t n, size_t m) { size_t i =...
s = set('spam') s = set(['a', 'b']) 特点: 无重复 操作: 查询: a in s (返回a是否存在于s集合) s|t s.union(t)(返回s和t并集) s&t s.intersection(t)(返回s和t交集) s-t s.difference(t)(返回s中有t没有的元素集合)