Question

我有一个字符串列表列表[每个元素都在括号中]，需要获取字符串列表，每个字符串表示每个列表中一个元素的元素组合。需要获得所有组合

[+9, +4, +a]
[+i*o -k*z +(m+n+f+5)][+i*o +(m+n+f+5) -k*z][+(m+n+f+5) +i*o -k*z]
[+h*i/o +6*l/8]
[+b/c/r +(a*b*x*y+1)][+(a*b*x*y+1)+b/c/r]

需要获得

[+9, +4, +a][+i*o -k*z +(m+n+f+5)][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+i*o -k*z +(m+n+f+5)][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]
[+9, +4, +a][+i*o +(m+n+f+5) -k*z][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+i*o +(m+n+f+5) -k*z][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]
[+9, +4, +a][+(m+n+f+5) +i*o -k*z][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+(m+n+f+5) +i*o -k*z][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]

任何帮助表示赞赏感谢

Answer 1

使用番石榴Sets.cartesianProduct：

import com.google.common.base.Joiner;
import com.google.common.collect.ImmutableSet;
import com.google.common.collect.Sets;

import java.util.List;
import java.util.Set;


public class Foo {
    public static void main(String[] args)
    {
        final Set<String> s1 = ImmutableSet.of("[+9, +4, +a]");
        final Set<String> s2 = ImmutableSet.of("[+i*o -k*z +(m+n+f+5)]","[+i*o +(m+n+f+5) -k*z]","[+(m+n+f+5) +i*o -k*z]");
        final Set<String> s3 = ImmutableSet.of("[+h*i/o +6*l/8]");
        final Set<String> s4 = ImmutableSet.of("[+b/c/r +(a*b*x*y+1)]","[+(a*b*x*y+1)+b/c/r]");
        @SuppressWarnings("unchecked")
        final Set<List<String>> cartesianProducts = Sets.cartesianProduct(s1, s2, s3, s4);
        for (final List<String> cartesianProduct : cartesianProducts) {
            System.out.println(Joiner.on("").join(cartesianProduct));
        }
    }        
}

输出：

[+9, +4, +a][+i*o -k*z +(m+n+f+5)][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+i*o +(m+n+f+5) -k*z][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+(m+n+f+5) +i*o -k*z][+h*i/o +6*l/8][+b/c/r +(a*b*x*y+1)]
[+9, +4, +a][+i*o -k*z +(m+n+f+5)][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]
[+9, +4, +a][+i*o +(m+n+f+5) -k*z][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]
[+9, +4, +a][+(m+n+f+5) +i*o -k*z][+h*i/o +6*l/8][+(a*b*x*y+1)+b/c/r]

Answer 2

我将如何做到这一点。这是未经测试的。

List<List<Object>> lists = ...;
int max = combinationCount(lists);
for (int i = 0; i < max; ++i)
{
    System.out.println(combination(lists, i));
}

使用：

public static int combinationCount(List<List> l)
{
   int r = 1;
   for (List i : l) r *= i.size();
   return r;
}

和

public static List combination(List<List> l, int index)
{
    List r = new ArrayList(l.size());
    for (int i = 0; i < l.size(); ++i)
    {
        List k = l.get(i);
        int m = index % k.size();
        index /= k.size();
        r.add(k.get(m));
    }
    return r;
}

合并n个列表保存顺序（Java）

2 个答案: