permutation

问题 void permute(string elems, int mid, int end) { static int count; if (mid == end) { cout << ++count << " : " << elems << endl; return ; } else { for (int i = mid; i <= end; i++) { swap(elems, mid, i); permute(elems, mid + 1, end); swap(elems, mid, i); } } } The above function shows the permutations of str (with str[0..mid-1] as a steady prefix, and str[mid..end] as a permutable suffix). So we can use permute(str, 0, str.size() - 1) to show all the permutations of one string. But the function

问题 I am implementing the following problem in ruby. Here's the pattern that I want : 1234, 1324, 1432, 1423, 2341 and so on i.e. the digits in the four digit number should be between [1-4] and should also be non-repetitive. to make you understand in a simple manner I take a two digit pattern and the solution should be : 12, 21 i.e. the digits should be either 1 or 2 and should be non-repetitive. To make sure that they are non-repetitive I want to use $1 for the condition for my second digit but

问题 I am implementing the following problem in ruby. Here's the pattern that I want : 1234, 1324, 1432, 1423, 2341 and so on i.e. the digits in the four digit number should be between [1-4] and should also be non-repetitive. to make you understand in a simple manner I take a two digit pattern and the solution should be : 12, 21 i.e. the digits should be either 1 or 2 and should be non-repetitive. To make sure that they are non-repetitive I want to use $1 for the condition for my second digit but

问题 I want to shuffle a list of unique items, but not do an entirely random shuffle. I need to be sure that no element in the shuffled list is at the same position as in the original list. Thus, if the original list is (A, B, C, D, E), this result would be OK: (C, D, B, E, A), but this one would not: (C, E, A, D, B) because "D" is still the fourth item. The list will have at most seven items. Extreme efficiency is not a consideration. I think this modification to Fisher/Yates does the trick, but

问题 I have an ArrayList[] myList and I am trying to create a list of all the permutations of the values in the arrays. EXAMPLE: (all values are strings) myList[0] = { "1", "5", "3", "9" }; myList[1] = { "2", "3" }; myList[2] = { "93" }; The count of myList can be varied so its length is not known beforehand. I would like to be able to generate a list of all the permutations similar to the following (but with some additional formatting). 1 2 93 1 3 93 5 2 93 5 3 93 3 2 93 3 3 93 9 2 93 9 3 93 Does

问题 I have an ArrayList[] myList and I am trying to create a list of all the permutations of the values in the arrays. EXAMPLE: (all values are strings) myList[0] = { "1", "5", "3", "9" }; myList[1] = { "2", "3" }; myList[2] = { "93" }; The count of myList can be varied so its length is not known beforehand. I would like to be able to generate a list of all the permutations similar to the following (but with some additional formatting). 1 2 93 1 3 93 5 2 93 5 3 93 3 2 93 3 3 93 9 2 93 9 3 93 Does

问题 I would like to get all combination of a number without any repetition. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme, but couldn't. I drew a graph/tree for it and this screams to use recursion. But I would like to do this without recursion, if this is possible. Can anyone please help me to do that? 回答1: Here is a generic permutation enumerator I wrote a year ago. It can also produce "sub-permutations": public class PermUtil <T> { private T[] arr; private int[]

问题 I would like to get all combination of a number without any repetition. Like 0.1.2, 0.2.1, 1.2.0, 1.0.2, 2.0.1, 2.1.0. I tried to find an easy scheme, but couldn't. I drew a graph/tree for it and this screams to use recursion. But I would like to do this without recursion, if this is possible. Can anyone please help me to do that? 回答1: Here is a generic permutation enumerator I wrote a year ago. It can also produce "sub-permutations": public class PermUtil <T> { private T[] arr; private int[]

问题 How can I randomly shuffle a list so that none of the elements remains in its original position? In other words, given a list A with distinct elements, I'd like to generate a permutation B of it so that this permutation is random and for each n , a[n] != b[n] e.g. a = [1,2,3,4] b = [4,1,2,3] # good b = [4,2,1,3] # good a = [1,2,3,4] x = [2,4,3,1] # bad I don't know the proper term for such a permutation (is it "total"?) thus having a hard time googling. The correct term appears to be

问题 I was curious how std:next_permutation was implemented so I extracted the the gnu libstdc++ 4.7 version and sanitized the identifiers and formatting to produce the following demo... #include <vector> #include <iostream> #include <algorithm> using namespace std; template<typename It> bool next_permutation(It begin, It end) { if (begin == end) return false; It i = begin; ++i; if (i == end) return false; i = end; --i; while (true) { It j = i; --i; if (*i < *j) { It k = end; while (!(*i < *--k))