英文:
How to add array values after exceeding a certain length in java
问题
我正在努力找出在数组值超过一定长度时添加这些值的最佳方法。该程序将允许用户将值输入数组中。一旦用户输入了数字的数量（可以是任意数量），这些数字不应超过6列，如果超过6列，则应从数组的右侧向左侧添加。请参见以下示例：
用户输入的12个数字：
arrNumbers = new int[] { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
应创建一个新数组，其内容如下所示：
newArrNumbers = int[] { 2, 4, 6, 8, 10, 12};
从而仅保留了6个插槽。
我不知道如何编写这段代码。我只知道可以使用类似下面的if语句来实现。我是一个初学者，希望从有经验的人那里了解解决方案。
if (arrNumbers.length > 6)
英文:
I am trying to work out the best way to add array values when they exceed a certain length. The program will allow a user to input values into the array. Once the user enters the amount of numbers (can be as much as they want). The numbers should not exceed 6 columns, and if they do they should be added from right to left of the array. See below
The 12 numbers inputted by the user:
arrNumbers = new int[] { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
A new array should be made which will look like this:
newArrNumbers = int [2, 4, 6, 8, 10, 12]
Resulting in only 6 slots.
I have no idea on how to code this. I only know that this could be possible with an if statement like below. I am a beginner and would like to know a solution from someone experienced.
if (arrNumbers.length > 6)
答案1
得分: 1
你可以轻松迭代遍历数组的其余部分。
int sum = 0;
for (int i = 6; i < arrNumbers.length; i++) {
sum += arrNumbers[i];
}
arrNumbers[5] = sum;
英文:
You can easily iterate through the rest of Array.
int sum=0;
for(int i=6;i<arrNumbers.length;i++){
sum+=arrNumbers[i];
}
arrNumbers[5]=sum;
</details>
# 答案2
**得分**: 1
我认为你正在寻找的是这个：
``` java
int[] arrNumbers = new int[] {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
int[] res = new int[6];
// 遍历初始数组并进行分块。根据要求，每块为6个元素。
for (int i = 0; i < arrNumbers.length; i += 6) {
// 创建包含前6个数字的子数组。
int[] ints = Arrays.copyOfRange(arrNumbers, i, Math.min(arrNumbers.length, i + 6));
// 将第一个块添加到结果数组中。重复直到完成。
for (int j = 0; j < ints.length; j++) {
res[j] += ints[j];
}
}
结果：res: {2, 4, 6, 8, 10, 12}
这适用于arrNumbers变量的任何数组大小。
英文:
I think what you are looking for is this
int[] arrNumbers = new int[] {1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
int[] res = new int[6];
// iterate over the initial array and chunk it. 6 will be our chunk per requirements.
for (int i = 0; i < arrNumbers.length; i += 6) {
// create a sub array for the first 6 numbers.
int[] ints = Arrays.copyOfRange(arrNumbers, i, Math.min(arrNumbers.length, i + 6));
// add the first chunk to our resulting array. Repeat until needed.
for (int j = 0; j < ints.length; j++) {
res[j] += ints[j];
}
}
Result: res: {2, 4, 6, 8, 10, 12}
This will work for any array size for the arrNumbers variable.
答案3
得分: 1
一个更加简洁高效的答案：
int[] arrNumbers = new int[] { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
int[] newArrNumbers = new int[6];
for(int i = 0; i < arrNumbers.length; i++){
newArrNumbers[i % 6] += arrNumbers[i];
}
输出: `[2, 4, 6, 8, 10, 12]`
英文:
A more concise and efficient answer:
int[] arrNumbers = new int[] { 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6};
int[] newArrNumbers = new int[6];
for(int i = 0; i < arrNumbers.length ; i++){
newArrNumbers[i % 6] += arrNumbers[i];
}
output: [2, 4, 6, 8, 10, 12]
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