Question

我试图从列表中找到所有可能的总和。该列表由用户输入的整数组成，输入数量由用户决定（参见代码）。我想要的是在没有目标的情况下检查所有可能的总和。我不想要目标的原因是因为程序随后被调整以选择最接近1000的这些总和中的一个。因此，如果我选择1000作为目标，我将无法说出1001。

来自用户的示例输入：

5 // user chose 5 numbers.

500,400,300,50,60  // numbers chosen by user.

输出将是：

1010 // because 500+400+60+50= 1010 and closest to 1000.

下一个例子可能是：

3 // user chose 3 numbers.

1,2,3 // numbers chosen by user.

输出将是：

6 // because 1+2+3 = 6.

回到我原来的问题，这是怎么做到的？每次我搜索一下所有可能的整数列表＆＃34;或类似于我得到一个目标，它在这个例子中起作用。

public static void main(String[] args) throws Exception {
    int bästaVikt;
    int räknare = 0;
    Scanner sc = new Scanner(System.in);
    ArrayList<Integer> mylist = new ArrayList<Integer>();
    int s;
    s = sc.nextInt();
    int ans = 0;
    for(int i = 1; i <= s; i++) {
        mylist.add(sc.nextInt());
    }
    for(int i = 0; i < mylist.size(); i++) {   
        Collections.sort(mylist);
        System.out.print(mylist.get(i));
        System.out.print(" ");  
    }   
}

Answer 1

半实际代码半伪代码，希望它有所帮助。

input // the array that holds all numbers than the user input
int n = input.length();
ArrayList<Integer> combinations = new ArrayList<Integer>();
int totalSum = input.sum(); // sum of all the values the user input
ArrayList<Integer> allSums = new ArrayList<Ingeter>();

for(int i = 0; i < n; i++){
    combinations = combine(i) // find all possible combinations of i integers in the input array; you can surely find code for retrieving combinations in another thread
    foreach combination in combinations{
        allSums.add(totalSum - combination.sum())
    }
    combinations = new ArrayList<Integer>(); // clear the combinations arraylist
}
return allSums; //all possible sums


minDistanceToNumber = absolute(allSums.get(0) - 1000); // define a function "int absolute(int value)" which will multiply value by -1 if it is less than 0 and return it or just return it if it is bigger than 0
minNumber = allSums.get(0); // this will hold the number which has the minimum distance to 1000, as in the example above
for each sum in allSums
    if(absolute(sum - 1000) < minDistanceToNumber){
        minDistanceToNumber = absolute(sum - 1000);
        minNumber = sum;
    }
}
return minNumber;

找到没有目标的列表的所有可能总和

1 个答案: