如何查找從唯一數組索引中繪製的數組中最大可能的數字總和

Question

我需要查找數組中最大可能的數字總和，但數字必須從唯一數組索引中繪製...就像那樣：

double maxSum = 0; 

double a = {1.0 , 2.0, 3.0}; 
double b = {4.0 , 5.0, 6.0}; 
double c = {7.0 , 8.0, 9.0}; 

sum = a[0] + b[1] + c[2]; 

// or sum = a[0] + b[2] + c[1] 
// or sum = a[1] + b[0] + c[2] 
// etc. - how to do that for i arrays and for j numbers in array? 

if(sum >=maxSum){ 
maxSum = sum; 
} 

這是我做過什麼 - 但沒有任何線索，下一步怎麼辦？

public ArrayList<Double[]> tempArrayCreator() { 
    ArrayList<Double[]> lists = new ArrayList<Double[]>(); 

    Double[] list1 = { 6.0, 7.0, 6.0 }; 
    Double[] list2 = { 4.5, 6.0, 6.75 }; 
    Double[] list3 = { 6.0, 5.0, 9.0 }; 

    lists.add(list1); 
    lists.add(list2); 
    lists.add(list3); 

    return lists; 
} 

public double maxPossibleSum(ArrayList<Double[]> lists) { 

    double result = 0.0; 

    for (int i = 0; i < lists.size(); i++) { 
     for (int j = 0; j < lists.get(i).length; j++) { 

     // ??? 

     } 
    } 

    return result; 

} 

編輯。例如：

list1 = { 1, 5, 2}; 
list2 = { 9, 3, 7}; 
list3 = { 8, 4, 9}; 

possible solutions: 
list1[0] + list2[1] + list3[2] = 13 
list1[0] + list2[2] + list3[1] = 12 
list1[1] + list2[0] + list3[2] = 23 <-- here it is! 
list1[1] + list2[2] + list3[0] = 20 
list1[2] + list2[0] + list3[1] = 15 
list1[2] + list2[1] + list3[0] = 13 

Answer 1

首先，你希望得到您的索引的所有排列的列表。

{[0, 1, 2], [0, 2, 1], [1, 0, 2], [1, 2, 0], [2, 1, 0], [2, 0, 1]} 

例如，這answer提供了一種方式來做到這一點：

public static List<List<Integer>> getAllPermutations(int arraySize) { 
    List<Integer> elements = new ArrayList<Integer>(); 
    for (int i = 0; i < arraySize; i++) { 
     elements.add(i); 
    } 
    List<List<Integer>> result = new ArrayList<List<Integer>>(); 
    getAllPermutations(result, elements, 0); 
    return result; 
} 

private static void getAllPermutations(List<List<Integer>> result, List<Integer> elements, int k) { 
    for (int i = k; i < elements.size(); i++) { 
     java.util.Collections.swap(elements, i, k); 
     getAllPermutations(result, elements, k + 1); 
     java.util.Collections.swap(elements, k, i); 
    } 
    if (k == elements.size() - 1) { 
     result.add(new ArrayList<Integer>(elements)); 
    } 
} 

然後你通過所有的排列組合循環：

public double maxPossibleSum(ArrayList<Double[]> lists) { 
    List<List<Integer>> allPermutations = getAllPermutations(3); 
    double maxSum = Double.NEGATIVE_INFINITY; 
    for (List<Integer> permutation : allPermutations) { 
     double sum = 0; 
     for (int i = 0; i < permutation.size(); i++) { 
      Integer index = permutation.get(i); 
      sum += lists.get(i)[index]; 
     } 
     if (sum > maxSum) { 
      maxSum = sum; 
     } 
    } 
    return maxSum; 
} 

Answer 2

for (int i = 0; i < lists.size(); i++) { 
     Double[] tmp = list.get(i); 
     for (int j = 0; j < tmp.length; j++) { 
      result += tmp[j];  
     }   

    } 

Answer 3

內的for循環只需添加

result+=lists.get(i)[j];