我想写一个函数，它以一个字母数组作为参数，并从中选择一些字母。
假设您提供了一个包含8个字母的数组，并希望从中选择3个字母。那么你应该得到:
8! / ((8 - 3)! * 3!) = 56
数组(或单词)，每个数组由3个字母组成。
当前回答
我在c++中为组合创建了一个通用类。 它是这样使用的。
char ar[] = "0ABCDEFGH";
nCr ncr(8, 3);
while(ncr.next()) {
for(int i=0; i<ncr.size(); i++) cout << ar[ncr[i]];
cout << ' ';
}
我的库ncr[i]从1返回，而不是从0返回。 这就是为什么数组中有0。 如果你想考虑订单，只需将nCr class改为nPr即可。 用法是相同的。
结果
美国广播公司 ABD 安倍 沛富 ABG ABH 澳洲牧牛犬 王牌 ACF ACG 呵呀 正面 ADF ADG 抗利尿激素 时 AEG AEH 二自由度陀螺仪 AFH 啊 BCD 公元前 供应量 波士顿咨询公司 BCH 12 快速公车提供 BDG BDH 性能试验 求 本· 高炉煤气 BFH 使用BGH CDE 提供 CDG 鼎晖 欧共体语言教学大纲的 CEG 另一 CFG CFH 全息 DEF 度 电气设施 脱硫 干扰 DGH EFG EFH EGH FGH
下面是头文件。
#pragma once
#include <exception>
class NRexception : public std::exception
{
public:
virtual const char* what() const throw() {
return "Combination : N, R should be positive integer!!";
}
};
class Combination
{
public:
Combination(int n, int r);
virtual ~Combination() { delete [] ar;}
int& operator[](unsigned i) {return ar[i];}
bool next();
int size() {return r;}
static int factorial(int n);
protected:
int* ar;
int n, r;
};
class nCr : public Combination
{
public:
nCr(int n, int r);
bool next();
int count() const;
};
class nTr : public Combination
{
public:
nTr(int n, int r);
bool next();
int count() const;
};
class nHr : public nTr
{
public:
nHr(int n, int r) : nTr(n,r) {}
bool next();
int count() const;
};
class nPr : public Combination
{
public:
nPr(int n, int r);
virtual ~nPr() {delete [] on;}
bool next();
void rewind();
int count() const;
private:
bool* on;
void inc_ar(int i);
};
以及执行。
#include "combi.h"
#include <set>
#include<cmath>
Combination::Combination(int n, int r)
{
//if(n < 1 || r < 1) throw NRexception();
ar = new int[r];
this->n = n;
this->r = r;
}
int Combination::factorial(int n)
{
return n == 1 ? n : n * factorial(n-1);
}
int nPr::count() const
{
return factorial(n)/factorial(n-r);
}
int nCr::count() const
{
return factorial(n)/factorial(n-r)/factorial(r);
}
int nTr::count() const
{
return pow(n, r);
}
int nHr::count() const
{
return factorial(n+r-1)/factorial(n-1)/factorial(r);
}
nCr::nCr(int n, int r) : Combination(n, r)
{
if(r == 0) return;
for(int i=0; i<r-1; i++) ar[i] = i + 1;
ar[r-1] = r-1;
}
nTr::nTr(int n, int r) : Combination(n, r)
{
for(int i=0; i<r-1; i++) ar[i] = 1;
ar[r-1] = 0;
}
bool nCr::next()
{
if(r == 0) return false;
ar[r-1]++;
int i = r-1;
while(ar[i] == n-r+2+i) {
if(--i == -1) return false;
ar[i]++;
}
while(i < r-1) ar[i+1] = ar[i++] + 1;
return true;
}
bool nTr::next()
{
ar[r-1]++;
int i = r-1;
while(ar[i] == n+1) {
ar[i] = 1;
if(--i == -1) return false;
ar[i]++;
}
return true;
}
bool nHr::next()
{
ar[r-1]++;
int i = r-1;
while(ar[i] == n+1) {
if(--i == -1) return false;
ar[i]++;
}
while(i < r-1) ar[i+1] = ar[i++];
return true;
}
nPr::nPr(int n, int r) : Combination(n, r)
{
on = new bool[n+2];
for(int i=0; i<n+2; i++) on[i] = false;
for(int i=0; i<r; i++) {
ar[i] = i + 1;
on[i] = true;
}
ar[r-1] = 0;
}
void nPr::rewind()
{
for(int i=0; i<r; i++) {
ar[i] = i + 1;
on[i] = true;
}
ar[r-1] = 0;
}
bool nPr::next()
{
inc_ar(r-1);
int i = r-1;
while(ar[i] == n+1) {
if(--i == -1) return false;
inc_ar(i);
}
while(i < r-1) {
ar[++i] = 0;
inc_ar(i);
}
return true;
}
void nPr::inc_ar(int i)
{
on[ar[i]] = false;
while(on[++ar[i]]);
if(ar[i] != n+1) on[ar[i]] = true;
}
其他回答
我有一个用于project euler的排列算法，用python编写:
def missing(miss,src):
"Returns the list of items in src not present in miss"
return [i for i in src if i not in miss]
def permutation_gen(n,l):
"Generates all the permutations of n items of the l list"
for i in l:
if n<=1: yield [i]
r = [i]
for j in permutation_gen(n-1,missing([i],l)): yield r+j
If
n<len(l)
你应该有所有你需要的组合，没有重复，你需要吗?
它是一个生成器，所以你可以这样使用它:
for comb in permutation_gen(3,list("ABCDEFGH")):
print comb
我知道这个问题已经有很多答案了，但我想在JavaScript中添加我自己的贡献，它由两个函数组成——一个生成原始n元素集的所有可能不同的k子集，另一个使用第一个函数生成原始n元素集的幂集。
下面是这两个函数的代码:
//Generate combination subsets from a base set of elements (passed as an array). This function should generate an
//array containing nCr elements, where nCr = n!/[r! (n-r)!].
//Arguments:
//[1] baseSet : The base set to create the subsets from (e.g., ["a", "b", "c", "d", "e", "f"])
//[2] cnt : The number of elements each subset is to contain (e.g., 3)
function MakeCombinationSubsets(baseSet, cnt)
{
var bLen = baseSet.length;
var indices = [];
var subSet = [];
var done = false;
var result = []; //Contains all the combination subsets generated
var done = false;
var i = 0;
var idx = 0;
var tmpIdx = 0;
var incr = 0;
var test = 0;
var newIndex = 0;
var inBounds = false;
var tmpIndices = [];
var checkBounds = false;
//First, generate an array whose elements are indices into the base set ...
for (i=0; i<cnt; i++)
indices.push(i);
//Now create a clone of this array, to be used in the loop itself ...
tmpIndices = [];
tmpIndices = tmpIndices.concat(indices);
//Now initialise the loop ...
idx = cnt - 1; //point to the last element of the indices array
incr = 0;
done = false;
while (!done)
{
//Create the current subset ...
subSet = []; //Make sure we begin with a completely empty subset before continuing ...
for (i=0; i<cnt; i++)
subSet.push(baseSet[tmpIndices[i]]); //Create the current subset, using items selected from the
//base set, using the indices array (which will change as we
//continue scanning) ...
//Add the subset thus created to the result set ...
result.push(subSet);
//Now update the indices used to select the elements of the subset. At the start, idx will point to the
//rightmost index in the indices array, but the moment that index moves out of bounds with respect to the
//base set, attention will be shifted to the next left index.
test = tmpIndices[idx] + 1;
if (test >= bLen)
{
//Here, we're about to move out of bounds with respect to the base set. We therefore need to scan back,
//and update indices to the left of the current one. Find the leftmost index in the indices array that
//isn't going to move out of bounds with respect to the base set ...
tmpIdx = idx - 1;
incr = 1;
inBounds = false; //Assume at start that the index we're checking in the loop below is out of bounds
checkBounds = true;
while (checkBounds)
{
if (tmpIdx < 0)
{
checkBounds = false; //Exit immediately at this point
}
else
{
newIndex = tmpIndices[tmpIdx] + 1;
test = newIndex + incr;
if (test >= bLen)
{
//Here, incrementing the current selected index will take that index out of bounds, so
//we move on to the next index to the left ...
tmpIdx--;
incr++;
}
else
{
//Here, the index will remain in bounds if we increment it, so we
//exit the loop and signal that we're in bounds ...
inBounds = true;
checkBounds = false;
//End if/else
}
//End if
}
//End while
}
//At this point, if we'er still in bounds, then we continue generating subsets, but if not, we abort immediately.
if (!inBounds)
done = true;
else
{
//Here, we're still in bounds. We need to update the indices accordingly. NOTE: at this point, although a
//left positioned index in the indices array may still be in bounds, incrementing it to generate indices to
//the right may take those indices out of bounds. We therefore need to check this as we perform the index
//updating of the indices array.
tmpIndices[tmpIdx] = newIndex;
inBounds = true;
checking = true;
i = tmpIdx + 1;
while (checking)
{
test = tmpIndices[i - 1] + 1; //Find out if incrementing the left adjacent index takes it out of bounds
if (test >= bLen)
{
inBounds = false; //If we move out of bounds, exit NOW ...
checking = false;
}
else
{
tmpIndices[i] = test; //Otherwise, update the indices array ...
i++; //Now move on to the next index to the right in the indices array ...
checking = (i < cnt); //And continue until we've exhausted all the indices array elements ...
//End if/else
}
//End while
}
//At this point, if the above updating of the indices array has moved any of its elements out of bounds,
//we abort subset construction from this point ...
if (!inBounds)
done = true;
//End if/else
}
}
else
{
//Here, the rightmost index under consideration isn't moving out of bounds with respect to the base set when
//we increment it, so we simply increment and continue the loop ...
tmpIndices[idx] = test;
//End if
}
//End while
}
return(result);
//End function
}
function MakePowerSet(baseSet)
{
var bLen = baseSet.length;
var result = [];
var i = 0;
var partialSet = [];
result.push([]); //add the empty set to the power set
for (i=1; i<bLen; i++)
{
partialSet = MakeCombinationSubsets(baseSet, i);
result = result.concat(partialSet);
//End i loop
}
//Now, finally, add the base set itself to the power set to make it complete ...
partialSet = [];
partialSet.push(baseSet);
result = result.concat(partialSet);
return(result);
//End function
}
我用集合["a"， "b"， "c"， "d"， "e"， "f"]作为基本集进行了测试，并运行代码以产生以下幂集:
[]
["a"]
["b"]
["c"]
["d"]
["e"]
["f"]
["a","b"]
["a","c"]
["a","d"]
["a","e"]
["a","f"]
["b","c"]
["b","d"]
["b","e"]
["b","f"]
["c","d"]
["c","e"]
["c","f"]
["d","e"]
["d","f"]
["e","f"]
["a","b","c"]
["a","b","d"]
["a","b","e"]
["a","b","f"]
["a","c","d"]
["a","c","e"]
["a","c","f"]
["a","d","e"]
["a","d","f"]
["a","e","f"]
["b","c","d"]
["b","c","e"]
["b","c","f"]
["b","d","e"]
["b","d","f"]
["b","e","f"]
["c","d","e"]
["c","d","f"]
["c","e","f"]
["d","e","f"]
["a","b","c","d"]
["a","b","c","e"]
["a","b","c","f"]
["a","b","d","e"]
["a","b","d","f"]
["a","b","e","f"]
["a","c","d","e"]
["a","c","d","f"]
["a","c","e","f"]
["a","d","e","f"]
["b","c","d","e"]
["b","c","d","f"]
["b","c","e","f"]
["b","d","e","f"]
["c","d","e","f"]
["a","b","c","d","e"]
["a","b","c","d","f"]
["a","b","c","e","f"]
["a","b","d","e","f"]
["a","c","d","e","f"]
["b","c","d","e","f"]
["a","b","c","d","e","f"]
只要复制粘贴这两个函数“原样”，你就有了提取n元素集的不同k子集所需的基本知识，并生成该n元素集的幂集(如果你愿意的话)。
我并不是说这很优雅，只是说它在经过大量的测试(并在调试阶段将空气变为蓝色:)之后可以工作。
这是我用c++写的命题
我尽可能少地限制迭代器类型，所以这个解决方案假设只有前向迭代器，它可以是const_iterator。这应该适用于任何标准容器。在参数没有意义的情况下，它抛出std:: invalid_argument
#include <vector>
#include <stdexcept>
template <typename Fci> // Fci - forward const iterator
std::vector<std::vector<Fci> >
enumerate_combinations(Fci begin, Fci end, unsigned int combination_size)
{
if(begin == end && combination_size > 0u)
throw std::invalid_argument("empty set and positive combination size!");
std::vector<std::vector<Fci> > result; // empty set of combinations
if(combination_size == 0u) return result; // there is exactly one combination of
// size 0 - emty set
std::vector<Fci> current_combination;
current_combination.reserve(combination_size + 1u); // I reserve one aditional slot
// in my vector to store
// the end sentinel there.
// The code is cleaner thanks to that
for(unsigned int i = 0u; i < combination_size && begin != end; ++i, ++begin)
{
current_combination.push_back(begin); // Construction of the first combination
}
// Since I assume the itarators support only incrementing, I have to iterate over
// the set to get its size, which is expensive. Here I had to itrate anyway to
// produce the first cobination, so I use the loop to also check the size.
if(current_combination.size() < combination_size)
throw std::invalid_argument("combination size > set size!");
result.push_back(current_combination); // Store the first combination in the results set
current_combination.push_back(end); // Here I add mentioned earlier sentinel to
// simplyfy rest of the code. If I did it
// earlier, previous statement would get ugly.
while(true)
{
unsigned int i = combination_size;
Fci tmp; // Thanks to the sentinel I can find first
do // iterator to change, simply by scaning
{ // from right to left and looking for the
tmp = current_combination[--i]; // first "bubble". The fact, that it's
++tmp; // a forward iterator makes it ugly but I
} // can't help it.
while(i > 0u && tmp == current_combination[i + 1u]);
// Here is probably my most obfuscated expression.
// Loop above looks for a "bubble". If there is no "bubble", that means, that
// current_combination is the last combination, Expression in the if statement
// below evaluates to true and the function exits returning result.
// If the "bubble" is found however, the ststement below has a sideeffect of
// incrementing the first iterator to the left of the "bubble".
if(++current_combination[i] == current_combination[i + 1u])
return result;
// Rest of the code sets posiotons of the rest of the iterstors
// (if there are any), that are to the right of the incremented one,
// to form next combination
while(++i < combination_size)
{
current_combination[i] = current_combination[i - 1u];
++current_combination[i];
}
// Below is the ugly side of using the sentinel. Well it had to haave some
// disadvantage. Try without it.
result.push_back(std::vector<Fci>(current_combination.begin(),
current_combination.end() - 1));
}
}
下面是一个方法，它从一个随机长度的字符串中给出指定大小的所有组合。类似于昆玛斯的解，但适用于不同的输入和k。
代码可以更改为换行，即'dab'从输入'abcd' w k=3。
public void run(String data, int howMany){
choose(data, howMany, new StringBuffer(), 0);
}
//n choose k
private void choose(String data, int k, StringBuffer result, int startIndex){
if (result.length()==k){
System.out.println(result.toString());
return;
}
for (int i=startIndex; i<data.length(); i++){
result.append(data.charAt(i));
choose(data,k,result, i+1);
result.setLength(result.length()-1);
}
}
"abcde"的输出:
ABC abd ace ade BCD bce bde cde
我已经编写了一个类来处理处理二项式系数的常见函数，这是您的问题属于的问题类型。它执行以下任务:
Outputs all the K-indexes in a nice format for any N choose K to a file. The K-indexes can be substituted with more descriptive strings or letters. This method makes solving this type of problem quite trivial. Converts the K-indexes to the proper index of an entry in the sorted binomial coefficient table. This technique is much faster than older published techniques that rely on iteration. It does this by using a mathematical property inherent in Pascal's Triangle. My paper talks about this. I believe I am the first to discover and publish this technique, but I could be wrong. Converts the index in a sorted binomial coefficient table to the corresponding K-indexes. Uses Mark Dominus method to calculate the binomial coefficient, which is much less likely to overflow and works with larger numbers. The class is written in .NET C# and provides a way to manage the objects related to the problem (if any) by using a generic list. The constructor of this class takes a bool value called InitTable that when true will create a generic list to hold the objects to be managed. If this value is false, then it will not create the table. The table does not need to be created in order to perform the 4 above methods. Accessor methods are provided to access the table. There is an associated test class which shows how to use the class and its methods. It has been extensively tested with 2 cases and there are no known bugs.
要了解这个类并下载代码，请参见将二项式系数表化。
将这个类转换为c++应该不难。
