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。
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
Here's some simple code that prints all the C(n,m) combinations. It works by initializing and moving a set of array indices that point to next valid combination. The indices are initialized to point to the lowest m indices (lexicographically the smallest combination). Then on, starting with the m-th index, we try to move the indices forward. if an index has reached its limit, we try the previous index (all the way down to index 1). If we can move an index forward, then we reset all greater indices.
m=(rand()%n)+1; // m will vary from 1 to n
for (i=0;i<n;i++) a[i]=i+1;
// we want to print all possible C(n,m) combinations of selecting m objects out of n
printf("Printing C(%d,%d) possible combinations ...\n", n,m);
// This is an adhoc algo that keeps m pointers to the next valid combination
for (i=0;i<m;i++) p[i]=i; // the p[.] contain indices to the a vector whose elements constitute next combination
done=false;
while (!done)
{
// print combination
for (i=0;i<m;i++) printf("%2d ", a[p[i]]);
printf("\n");
// update combination
// method: start with p[m-1]. try to increment it. if it is already at the end, then try moving p[m-2] ahead.
// if this is possible, then reset p[m-1] to 1 more than (the new) p[m-2].
// if p[m-2] can not also be moved, then try p[m-3]. move that ahead. then reset p[m-2] and p[m-1].
// repeat all the way down to p[0]. if p[0] can not also be moved, then we have generated all combinations.
j=m-1;
i=1;
move_found=false;
while ((j>=0) && !move_found)
{
if (p[j]<(n-i))
{
move_found=true;
p[j]++; // point p[j] to next index
for (k=j+1;k<m;k++)
{
p[k]=p[j]+(k-j);
}
}
else
{
j--;
i++;
}
}
if (!move_found) done=true;
}
其他回答
简短的python代码,产生索引位置
def yield_combos(n,k):
# n is set size, k is combo size
i = 0
a = [0]*k
while i > -1:
for j in range(i+1, k):
a[j] = a[j-1]+1
i=j
yield a
while a[i] == i + n - k:
i -= 1
a[i] += 1
下面的递归算法从有序集中选取所有k元素组合:
选择组合中的第一个元素I 将I与从大于I的元素集中递归选择的k-1个元素的组合组合。
对集合中的每一个i进行上述迭代。
为了避免重复,您必须选择比i大的其余元素。这样[3,5]将只被选中一次,即[3]与[5]结合,而不是两次(该条件消除了[5]+[3])。没有这个条件,你得到的是变化而不是组合。
为此,我在SQL Server 2005中创建了一个解决方案,并将其发布在我的网站上:http://www.jessemclain.com/downloads/code/sql/fn_GetMChooseNCombos.sql.htm
下面是一个例子来说明用法:
SELECT * FROM dbo.fn_GetMChooseNCombos('ABCD', 2, '')
结果:
Word
----
AB
AC
AD
BC
BD
CD
(6 row(s) affected)
在VB。Net,该算法从一组数字(PoolArray)中收集n个数字的所有组合。例如,从“8,10,20,33,41,44,47”中选择5个选项的所有组合。
Sub CreateAllCombinationsOfPicksFromPool(ByVal PicksArray() As UInteger, ByVal PicksIndex As UInteger, ByVal PoolArray() As UInteger, ByVal PoolIndex As UInteger)
If PicksIndex < PicksArray.Length Then
For i As Integer = PoolIndex To PoolArray.Length - PicksArray.Length + PicksIndex
PicksArray(PicksIndex) = PoolArray(i)
CreateAllCombinationsOfPicksFromPool(PicksArray, PicksIndex + 1, PoolArray, i + 1)
Next
Else
' completed combination. build your collections using PicksArray.
End If
End Sub
Dim PoolArray() As UInteger = Array.ConvertAll("8,10,20,33,41,44,47".Split(","), Function(u) UInteger.Parse(u))
Dim nPicks as UInteger = 5
Dim Picks(nPicks - 1) As UInteger
CreateAllCombinationsOfPicksFromPool(Picks, 0, PoolArray, 0)
这是一个简单的JS解决方案:
function getAllCombinations(n, k, f1) { indexes = Array(k); for (let i =0; i< k; i++) { indexes[i] = i; } var total = 1; f1(indexes); while (indexes[0] !== n-k) { total++; getNext(n, indexes); f1(indexes); } return {total}; } function getNext(n, vec) { const k = vec.length; vec[k-1]++; for (var i=0; i<k; i++) { var currentIndex = k-i-1; if (vec[currentIndex] === n - i) { var nextIndex = k-i-2; vec[nextIndex]++; vec[currentIndex] = vec[nextIndex] + 1; } } for (var i=1; i<k; i++) { if (vec[i] === n - (k-i - 1)) { vec[i] = vec[i-1] + 1; } } return vec; } let start = new Date(); let result = getAllCombinations(10, 3, indexes => console.log(indexes)); let runTime = new Date() - start; console.log({ result, runTime });