英文:
How do you add all elements in an 2d array, circularly?
问题
给定一个二维字符串数组，如何将所有元素相加，使数组中的每个索引都是数组其余部分的和呢？例如，对于数组[[a], [b], [c], [d], [e]]，最终结果是
[[a+b+c+d+e], [a+b+c+d+e], [a+b+c+d+e], [a+b+c+d+e],[a+b+c+d+e]]。我猜不一定要是二维数组，也可以是一维字符串数组，最终每个索引都是字符串"a+b+c+d+e"。当我说要循环时，我是指使用循环，不使用额外的内存空间。因此，也许在第一次迭代之后，它可能会看起来像[[a], [a + b], [a+b+c], [a+b+c+d], [a+b+c+d+e]]。实现这样的代码是什么样的呢？答案可以是伪代码或任何编程语言。
类似这样的链接：https://andrew.gibiansky.com/blog/machine-learning/baidu-allreduce/
英文:
Given an 2d array of strings how to I add up all the elements so that each index in the array is the sum of the rest of the array? For example with an array [[a], [b], [c], [d], [e]] I end up with
[[a+b+c+d+e], [a+b+c+d+e], [a+b+c+d+e], [a+b+c+d+e],[a+b+c+d+e]]. I guess it doesn't have to be a 2d array it can be a 1d array of string and I end up with a string "a+b+c+d+e" in every index. When I say circularly I mean with loops, no extra memory space. So maybe after the first iteration it could look something like [[a], [a + b], [a+b+c], [a+b+c+d], [a+b+c+d+e]] . What would the code for something like this be? The answer can be is psuedocode or any language
something like this: https://andrew.gibiansky.com/blog/machine-learning/baidu-allreduce/
答案1
得分: 1
我不太确定您想要实现什么。但是，如果您查看Arrays#parallelPrefix，您可以使用它来累积数组的单个元素。
public static void main(String[] args) {
String[] str = {"a","b","c","d","e"};
Arrays.parallelPrefix(str, (s1,s2) -> s1+s2);
System.out.println(Arrays.toString(str));
}
//[a, ab, abc, abcd, abcde]
英文:
I'm not quite sure what you want to achieve. But it might be worthwhile if you look at Arrays#parallelPrefix , which you can use to cumulate the single elements of an array.
public static void main(String[] args) {
String[] str = {"a","b","c","d","e"};
Arrays.parallelPrefix(str, (s1,s2) -> s1+s2);
System.out.println(Arrays.toString(str));
}
//[a, ab, abc, abcd, abcde]
答案2
得分: 1
你可以通过将所有元素添加到其中一个项目中，然后将结果复制到其他项目中来获得这样的字符串。
String[] arr = new String[] {"a", "b", "c", "d", "e"};
for(int i = 1; i < arr.length; ++i) {
arr[0] += arr[i];
}
for(int i = 1; i < arr.length; ++i) {
arr[i] = arr[0];
}
System.out.println(Arrays.toString(arr));
英文:
You can get such string by adding all elements to one of the items and then copying the result to other items.
String[] arr = new String[] {"a", "b", "c", "d", "e"};
for(int i = 1; i < arr.length; ++i) {
arr[0] += arr[i];
}
for(int i = 1; i < arr.length; ++i) {
arr[i] = arr[0];
}
System.out.println(Arrays.toString(arr));
答案3
得分: 1
根据您附上的链接，我认为以下解决方案会对您有所帮助：
int n = 5, i, j;
String[][] gpu = new String[n][n];
// 在此生成初始 GPU 数组...
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
char ch = (char) (97 + j);
gpu[i][j] = "" + ch + i;
}
}
System.out.println("初始 GPU 数组:");
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
System.out.print(gpu[i][j] + "|");
}
System.out.println();
}
// 在此计算您期望的逻辑
for (i = 0; i < n-1; i++) {
for (j = 0; j < n; j++) {
gpu[(i+j+1)%n][j] = gpu[(i+j)%n][j] + gpu[(i+j+1)%n][j];
}
}
System.out.println("最终 GPU 数组:");
for (i = 0; i < n; i++) {
System.out.print(gpu[n-1][i] + "|");
}
System.out.println();
英文:
According to your attached link i think below solution will help you:
int n = 5, i, j;
String[][] gpu = new String[n][n];
//here generate initial gpu....
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
char ch = (char) (97 + j);
gpu[i][j] = ""+ch+i;
}
}
System.out.println("Initial Gpu: ");
for (i = 0; i < n; i++) {
for (j = 0; j < n; j++) {
System.out.print(gpu[i][j] + "|");
}
System.out.println();
}
// here calculated your expected logic
for (i = 0; i < n-1; i++) {
for (j = 0; j < n; j++) {
gpu[(i+j+1)%n][j] = gpu[(i+j)%n][j] + gpu[(i+j+1)%n][j];
}
}
System.out.println("Final Gpu: ");
for (i = 0; i < n; i++) {
System.out.print(gpu[n-1][i] + "|");
}
System.out.println();
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