lower-bound | 易学教程

understanding of lower bound for comparison-based sorting algorithm

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问题 First, I know lower bound is O(nlogn) and how to prove it And I agree the lower bound should be O(nlogn). What I don't quite understand is: For some special cases, the # of comparisons could actually be even lower than the lower bound. For example, use bubble sort to sort an already sorted array. The # of comparisons is O(n). So how to actually understand the idea of lower bound ? The classical definition on Wikipedial: http://en.wikipedia.org/wiki/Upper_and_lower_bounds does not help much.

Can I extend std::map::lower_bound to search on non-key_type arguments?

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问题 Here is an illustration of my situation. I have a std::map and I want to find the first pair<key,value> where the key is any member of an equivalence class of keys. #include <map> struct Category { int foo; int bar; bool operator < (const Category & rhs) const; bool operator > (const Category & rhs) const; }; struct Key { Category category; float quality; bool operator < (const Key & rhs) const { if (category < rhs.category) return true; else if (category > rhs.category) return false; else

Regarding complexity (if comparison based sorting algorithm used)

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问题 as we all know that any sorting algorithm based on comparison model has lower bound of nlogn i.e Omega(nlogn). which can be proved mathematically. but as we all know dutch flag problem can sort 3 distinct elements (used repeatedly) in O(n) time.It is also based on comparison model but can sort in O(n) time. so how can we justify lower bound of sorting based on comparison model , because dutch flag problem kinda violates that. please help me understanding the difference..... Thanks 回答1: In my

<algorithm> function for finding last item less-than-or-equal to, like lower_bound

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问题 Is there a function in that uses binary search, like lower_bound but that returns the last item less-than-or-equal-to according to a given predicate? lower_bound is defined to: Finds the position of the first element in an ordered range that has a value greater than or equivalent to a specified value, where the ordering criterion may be specified by a binary predicate. and upper_bound : Finds the position of the first element in an ordered range that has a value that is greater than a

rationale for std::lower_bound and std::upper_bound?

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问题 STL provides binary search functions std::lower_bound and std::upper_bound, but I tend not to use them because I've been unable to remember what they do, because their contracts seem completely mystifying to me. Just from looking at the names, I'd guess that "lower_bound" might be short for "last lower bound", i.e. the last element in the sorted list that is <= the given val (if any). And similarly I'd guess "upper_bound" might be short for "first upper bound", i.e. the first element in the

Upper/lower bounds don't work as I expect, can not understand why

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问题 Here is the code. As a result I get "4 4". Don't understand why it is not "2 4" (according to lower and upper bounds' defenitions). #include <bits/stdc++.h> using namespace std; int main() { vector<int> v = {1, 2, 4, 5}; vector<int>::iterator s , f; s = lower_bound(v.begin(), v.end(), 3); f = upper_bound(v.begin(), v.end(), 3); cout << (*s) << " " << (*f); return 0; } 回答1: From std::lower_bound: Returns an iterator pointing to the first element in the range [first,last) which does not compare

rationale for std::lower_bound and std::upper_bound?

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STL provides binary search functions std::lower_bound and std::upper_bound, but I tend not to use them because I've been unable to remember what they do, because their contracts seem completely mystifying to me. Just from looking at the names, I'd guess that "lower_bound" might be short for "last lower bound", i.e. the last element in the sorted list that is <= the given val (if any). And similarly I'd guess "upper_bound" might be short for "first upper bound", i.e. the first element in the sorted list that is >= the given val (if any). But the documentation says they do something rather

lower_bound for vector<MyClass*>

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问题 I have this simple class: class MyClass { public: int id; string name; }; I want to have a vector with pointers to objects of this class that is sorted by the referenced MyClass id . I thought that using lower_bound would be easy, I did it before with vectors of objects (not pointers). With objects, I overloaded operator< like that: bool operator<(MyClass left, int right) { return (left.id < right); } Then I used lower_bound to insert new MyClass object to sorted vector. vector<MyClass>:

lower_bound for vector<MyClass*>

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I have this simple class: class MyClass { public: int id; string name; }; I want to have a vector with pointers to objects of this class that is sorted by the referenced MyClass id . I thought that using lower_bound would be easy, I did it before with vectors of objects (not pointers). With objects, I overloaded operator< like that: bool operator<(MyClass left, int right) { return (left.id < right); } Then I used lower_bound to insert new MyClass object to sorted vector. vector<MyClass>::iterator low; low = lower_bound(vectorname.begin(),vectorname.end(),id); prP = idContainer.begin(); prP =

Difference between basic binary search for upper bound and lower bound?

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In the article http://community.topcoder.com/tc?module=Static&d1=tutorials&d2=binarySearch , the author discusses binary search. He makes a distinction between finding the lowest value where something is true, and the highest value where something is false. The array being searched looks something like: false false false true true I am curious as to why these two cases are different. Why can't you just find the lowest value which is true, then subtract one to find the highest value which is false? Edit2: Ok, so I understand lower vs upper bound. Now, I am struggling to understand, when