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
下面是我最近用Java写的一段代码，它计算并返回从“outOf”元素中“num”元素的所有组合。
// author: Sourabh Bhat (heySourabh@gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}
其他回答
这是一个c++解决方案，我提出使用递归和位移位。它也可以在C语言中工作。
void r_nCr(unsigned int startNum, unsigned int bitVal, unsigned int testNum) // Should be called with arguments (2^r)-1, 2^(r-1), 2^(n-1)
{
unsigned int n = (startNum - bitVal) << 1;
n += bitVal ? 1 : 0;
for (unsigned int i = log2(testNum) + 1; i > 0; i--) // Prints combination as a series of 1s and 0s
cout << (n >> (i - 1) & 1);
cout << endl;
if (!(n & testNum) && n != startNum)
r_nCr(n, bitVal, testNum);
if (bitVal && bitVal < testNum)
r_nCr(startNum, bitVal >> 1, testNum);
}
你可以在这里找到这是如何工作的解释。
也许我错过了重点(你需要的是算法，而不是现成的解决方案)，但看起来scala已经开箱即用了(现在):
def combis(str:String, k:Int):Array[String] = {
str.combinations(k).toArray
}
使用这样的方法:
println(combis("abcd",2).toList)
会产生:
List(ab, ac, ad, bc, bd, cd)
另一种python递归解决方案。
def combination_indicies(n, k, j = 0, stack = []):
if len(stack) == k:
yield list(stack)
return
for i in range(j, n):
stack.append(i)
for x in combination_indicies(n, k, i + 1, stack):
yield x
stack.pop()
list(combination_indicies(5, 3))
输出:
[[0, 1, 2],
[0, 1, 3],
[0, 1, 4],
[0, 2, 3],
[0, 2, 4],
[0, 3, 4],
[1, 2, 3],
[1, 2, 4],
[1, 3, 4],
[2, 3, 4]]
《计算机编程艺术，卷4A:组合算法，第1部分》第7.2.1.3节中算法L(字典组合)的C代码:
#include <stdio.h>
#include <stdlib.h>
void visit(int* c, int t)
{
// for (int j = 1; j <= t; j++)
for (int j = t; j > 0; j--)
printf("%d ", c[j]);
printf("\n");
}
int* initialize(int n, int t)
{
// c[0] not used
int *c = (int*) malloc((t + 3) * sizeof(int));
for (int j = 1; j <= t; j++)
c[j] = j - 1;
c[t+1] = n;
c[t+2] = 0;
return c;
}
void comb(int n, int t)
{
int *c = initialize(n, t);
int j;
for (;;) {
visit(c, t);
j = 1;
while (c[j]+1 == c[j+1]) {
c[j] = j - 1;
++j;
}
if (j > t)
return;
++c[j];
}
free(c);
}
int main(int argc, char *argv[])
{
comb(5, 3);
return 0;
}
下面是我最近用Java写的一段代码，它计算并返回从“outOf”元素中“num”元素的所有组合。
// author: Sourabh Bhat (heySourabh@gmail.com)
public class Testing
{
public static void main(String[] args)
{
// Test case num = 5, outOf = 8.
int num = 5;
int outOf = 8;
int[][] combinations = getCombinations(num, outOf);
for (int i = 0; i < combinations.length; i++)
{
for (int j = 0; j < combinations[i].length; j++)
{
System.out.print(combinations[i][j] + " ");
}
System.out.println();
}
}
private static int[][] getCombinations(int num, int outOf)
{
int possibilities = get_nCr(outOf, num);
int[][] combinations = new int[possibilities][num];
int arrayPointer = 0;
int[] counter = new int[num];
for (int i = 0; i < num; i++)
{
counter[i] = i;
}
breakLoop: while (true)
{
// Initializing part
for (int i = 1; i < num; i++)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i] = counter[i - 1] + 1;
}
// Testing part
for (int i = 0; i < num; i++)
{
if (counter[i] < outOf)
{
continue;
} else
{
break breakLoop;
}
}
// Innermost part
combinations[arrayPointer] = counter.clone();
arrayPointer++;
// Incrementing part
counter[num - 1]++;
for (int i = num - 1; i >= 1; i--)
{
if (counter[i] >= outOf - (num - 1 - i))
counter[i - 1]++;
}
}
return combinations;
}
private static int get_nCr(int n, int r)
{
if(r > n)
{
throw new ArithmeticException("r is greater then n");
}
long numerator = 1;
long denominator = 1;
for (int i = n; i >= r + 1; i--)
{
numerator *= i;
}
for (int i = 2; i <= n - r; i++)
{
denominator *= i;
}
return (int) (numerator / denominator);
}
}
