Question

我有一个二维列表，其中行数在1到5之间，每行中的元素数是动态的。假设我有一个类似

的列表

values = [[1,2,3], [4,5],[6,7,8,9]];

result = [[1,4,6],[1,4,7],[1,4,8],[1,4,9],[1,5,6],[1,5,7],[1,5,8],[1,5,9],[2,4,6],[2,4,7],[2,4,8],[2,4,9],[2,5,6],[2,5,7],[2,5,8],[2,5,9],[3,4,6],[3,4,7],[3,4,8],[3,4,9],[3,5,6],[3,5,7],[3,5,8],[3,5,9]]

值是我的输入，我需要以结果格式显示输出。

我尝试以递归方式实现功能，但失败了。我的功能是

import groovy.transform.Field

List lines = [];
def function(row, col, lines)
{       
    if(row >= values.size())
    {           
        log.info("This should go inside result: " + lines);
        return;     
    }
    if(col >= values[row].size())
    {           
        return;         
    }   

    lines << values[row][col];  
    function(row+1,col,lines);
    lines.pop();
    function(row,col+1,lines);  
    return;
}


@Field
List values = [[1,2,3],[4,5],[6,7,8,9]];

@Field
List lst = [];


for(int i=0; i<values[0].size(); i++)
    function(0, i, lines)

Answer 1

Groovy具有内置的收集功能combinations()，可从元素列表中生成所有可能的组合。

用法示例：

assert [['a', 'b'],[1, 2, 3]].combinations() == [['a', 1], ['b', 1], ['a', 2], ['b', 2], ['a', 3], ['b', 3]]

您可以从初始脚本中简单地在values上调用此方法：

def values = [[1,2,3], [4,5],[6,7,8,9]]

def expected = [[1,4,6],[1,4,7],[1,4,8],[1,4,9],[1,5,6],[1,5,7],[1,5,8],[1,5,9],[2,4,6],[2,4,7],[2,4,8],[2,4,9],[2,5,6],[2,5,7],[2,5,8],[2,5,9],[3,4,6],[3,4,7],[3,4,8],[3,4,9],[3,5,6],[3,5,7],[3,5,8],[3,5,9]]

def result = values.combinations()

assert ((expected as Set) == (result as Set))

但是，如果您真的想自己实现此算法，则可以实现一种非常简单的尾部递归算法，如下所示：

import groovy.transform.CompileStatic
import groovy.transform.TailRecursive

@TailRecursive
@CompileStatic
<T> List<List<T>> combinations(final List<List<T>> xss, final List<List<T>> result = [[]]) {
    return !xss ? result : combinations(xss.tail(), process(xss.head(), result))
}

@CompileStatic
<T> List<List<T>> process(final List<T> xs, final List<List<T>> result) {
    return result.inject([]) { yss, ys -> yss + xs.collect { x -> ys + x } }
}

def values = [[1,2,3], [4,5],[6,7,8,9]]

def expected = [[1,4,6],[1,4,7],[1,4,8],[1,4,9],[1,5,6],[1,5,7],[1,5,8],[1,5,9],[2,4,6],[2,4,7],[2,4,8],[2,4,9],[2,5,6],[2,5,7],[2,5,8],[2,5,9],[3,4,6],[3,4,7],[3,4,8],[3,4,9],[3,5,6],[3,5,7],[3,5,8],[3,5,9]]

assert combinations(values) == expected

Answer 2

class Some {

    static main(args) {

        def values = [[1,2,3], [4,5],[6,7,8,9]];

        int[] index = new int[values.size()];
        index.eachWithIndex { v, i ->
            index[i] = 0
        }
        print "["
        recursive(index, values);
        print "]"
    }


    public static recursive(int[] index, def values){
        print "["
        (0..(index.length - 1) ).each { i -> 
            print values[i][index[i]] + ( i == index.length - 1 ? "" : ",") 
        }
        print "]"


        boolean allVisited = false;
        for(int i = index.length - 1; i >= 0; i--  ){
            if (index[i] < values[i].size() - 1){
                index[i]++;
                for(int j = i + 1; j < index.length; j++  ){
                        index[j] = 0;
                    }
                break;
            }else if (i == 0){
                allVisited = true;
                break;
            }
        }

        if (!allVisited){
            print ", "
            recursive(index, values);
        }
    }

}

这是递归算法。希望对您有所帮助。

在Groovy中生成2D列表元素的组合

2 个答案: