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
在Python中,利用递归的优势和所有事情都是通过引用完成的事实。对于非常大的集合,这将占用大量内存,但其优点是初始集合可以是一个复杂的对象。它只会找到唯一的组合。
import copy
def find_combinations( length, set, combinations = None, candidate = None ):
# recursive function to calculate all unique combinations of unique values
# from [set], given combinations of [length]. The result is populated
# into the 'combinations' list.
#
if combinations == None:
combinations = []
if candidate == None:
candidate = []
for item in set:
if item in candidate:
# this item already appears in the current combination somewhere.
# skip it
continue
attempt = copy.deepcopy(candidate)
attempt.append(item)
# sorting the subset is what gives us completely unique combinations,
# so that [1, 2, 3] and [1, 3, 2] will be treated as equals
attempt.sort()
if len(attempt) < length:
# the current attempt at finding a new combination is still too
# short, so add another item to the end of the set
# yay recursion!
find_combinations( length, set, combinations, attempt )
else:
# the current combination attempt is the right length. If it
# already appears in the list of found combinations then we'll
# skip it.
if attempt in combinations:
continue
else:
# otherwise, we append it to the list of found combinations
# and move on.
combinations.append(attempt)
continue
return len(combinations)
你可以这样使用它。传递'result'是可选的,所以你可以用它来获取可能组合的数量…尽管这样做效率很低(最好通过计算来完成)。
size = 3
set = [1, 2, 3, 4, 5]
result = []
num = find_combinations( size, set, result )
print "size %d results in %d sets" % (size, num)
print "result: %s" % (result,)
您应该从测试数据中得到以下输出:
size 3 results in 10 sets
result: [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 3, 4], [1, 3, 5], [1, 4, 5], [2, 3, 4], [2, 3, 5], [2, 4, 5], [3, 4, 5]]
如果你的集合是这样的,它也会工作得很好:
set = [
[ 'vanilla', 'cupcake' ],
[ 'chocolate', 'pudding' ],
[ 'vanilla', 'pudding' ],
[ 'chocolate', 'cookie' ],
[ 'mint', 'cookie' ]
]
其他回答
下面是我的JavaScript解决方案,通过使用reduce/map,它消除了几乎所有变量,功能更强大
function combinations(arr, size) { var len = arr.length; if (size > len) return []; if (!size) return [[]]; if (size == len) return [arr]; return arr.reduce(function (acc, val, i) { var res = combinations(arr.slice(i + 1), size - 1) .map(function (comb) { return [val].concat(comb); }); return acc.concat(res); }, []); } var combs = combinations([1,2,3,4,5,6,7,8],3); combs.map(function (comb) { document.body.innerHTML += comb.toString() + '<br />'; }); document.body.innerHTML += '<br /> Total combinations = ' + combs.length;
static IEnumerable<string> Combinations(List<string> characters, int length)
{
for (int i = 0; i < characters.Count; i++)
{
// only want 1 character, just return this one
if (length == 1)
yield return characters[i];
// want more than one character, return this one plus all combinations one shorter
// only use characters after the current one for the rest of the combinations
else
foreach (string next in Combinations(characters.GetRange(i + 1, characters.Count - (i + 1)), length - 1))
yield return characters[i] + next;
}
}
我可以给出这个问题的递归Python解决方案吗?
def choose_iter(elements, length):
for i in xrange(len(elements)):
if length == 1:
yield (elements[i],)
else:
for next in choose_iter(elements[i+1:], length-1):
yield (elements[i],) + next
def choose(l, k):
return list(choose_iter(l, k))
使用示例:
>>> len(list(choose_iter("abcdefgh",3)))
56
我喜欢它的简洁。
还有另一个递归解决方案(你应该能够使用字母而不是数字)使用堆栈,虽然比大多数更短:
stack = []
def choose(n,x):
r(0,0,n+1,x)
def r(p, c, n,x):
if x-c == 0:
print stack
return
for i in range(p, n-(x-1)+c):
stack.append(i)
r(i+1,c+1,n,x)
stack.pop()
4选3或者我想要从0到4的所有3种数字组合
choose(4,3)
[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]
假设你的字母数组是这样的:"ABCDEFGH"。你有三个下标(i, j, k)来表示你要用哪个字母来表示当前单词。
A B C D E F G H ^ ^ ^ i j k
首先你改变k,所以下一步看起来像这样:
A B C D E F G H ^ ^ ^ i j k
如果你到达终点,你继续改变j和k。
A B C D E F G H ^ ^ ^ i j k A B C D E F G H ^ ^ ^ i j k
一旦j达到G, i也开始变化。
A B C D E F G H ^ ^ ^ i j k A B C D E F G H ^ ^ ^ i j k ...
用代码写出来是这样的
void print_combinations(const char *string)
{
int i, j, k;
int len = strlen(string);
for (i = 0; i < len - 2; i++)
{
for (j = i + 1; j < len - 1; j++)
{
for (k = j + 1; k < len; k++)
printf("%c%c%c\n", string[i], string[j], string[k]);
}
}
}