英文:
How to get my function to be recursive? (Java)
问题
我正在编写一个递归函数，它将接收一个存有 "int" 元素的栈，并输出栈中元素平方和的结果。以下是我的代码：
public int sum_sqr_rec(Stack<Integer> stk) {
int sum = 0;
for (int i = 0; i < stk.size(); i++) {
sum += (stk.get(i) * stk.get(i));
}
return sum;
}
英文:
I'm having to make a recursive function that will receive a stack of "int" and output the sum of the squares of the elements in the stack.
Here is what I have
public int sum_sqr_rec(Stack<Integer> stk){
int sum = 0;
for (int i=0; i<stk.size(); i++){
sum += (stk.get(i) * stk.get(i));
}
return sum;
}
答案1
得分: 1
以下是翻译好的内容：
递归函数中最重要的是确定何时终止它。
第二个需要考虑的重要因素是在终止递归时返回什么。当你开始累加数字时，你可以从 sum = 0 开始。对于一个用于计算数字和的递归函数，在终止时可以返回相同的值（即 0）。同样地，对于一个用于计算数字乘积的递归函数，在终止时可以返回 1。
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Stack<Integer> stack = new Stack<Integer>();
stack.add(2);
stack.add(3);
stack.add(4);
stack.add(5);
System.out.println(sum_sqr_rec(stack));
}
static int sum_sqr_rec(Stack<Integer> stk) {
if (stk.isEmpty()) {
return 0;
}
int n = stk.pop();
return n * n + sum_sqr_rec(stk);
}
}
输出：
54
英文:
The most important thing you need to determine for a recursive function is when to terminate it.
The second important thing to consider is what to return when you terminate it. When you start adding numbers, you start with sum = 0. From a recursive function, which is supposed to calculate the sum of numbers, the same value (i.e. 0) can be returned when you terminate it. Similarly, from a recursive function, which is supposed to return the product of numbers, you can return 1 on termination.
import java.util.Stack;
public class Main {
public static void main(String[] args) {
Stack<Integer> stack = new Stack<Integer>();
stack.add(2);
stack.add(3);
stack.add(4);
stack.add(5);
System.out.println(sum_sqr_rec(stack));
}
static int sum_sqr_rec(Stack<Integer> stk) {
if (stk.isEmpty()) {
return 0;
}
int n = stk.pop();
return n * n + sum_sqr_rec(stk);
}
}
Output:
54
答案2
得分: 0
你可以像这样使用递归：
public static void main(String[] args) {
Stack<Integer> stack = new Stack<>();
stack.add(2);
stack.add(2);
stack.add(4);
stack.add(5);
System.out.println(sum_sqr_rec(stack));
}
public static int sum_sqr_rec(Stack<Integer> stack) {
if (stack.isEmpty())
return 0;
return stack.peek() * stack.pop() + sum_sqr_rec(stack);
}
英文:
You can use recursion like this:
public static void main(String[] args) {
Stack<Integer> stack = new Stack<>();
stack.add(2);
stack.add(2);
stack.add(4);
stack.add(5);
System.out.println(sum_sqr_rec(stack));
}
public static int sum_sqr_rec(Stack<Integer> stack) {
if (stack.isEmpty())
return 0;
return stack.peek() * stack.pop() + sum_sqr_rec(stack);
}
答案3
得分: 0
注意，我使用的是Deque接口，而不是直接使用Stack类（文档建议优先使用Deque接口）。
public int recursive(Deque<Integer> stk){
if (stk.isEmpty()){
return 0;
}
return (int) Math.pow(stk.pop(), 2) + recursive(stk);
}
英文:
Note that I use Deque interface rather than Stack class directly (documentation says it should be used in preference to the Stack class).
public int recursive(Deque<Integer> stk){
if (stk.Empty()){
return 0;
}
return Math.pow(stack.pop(), 2) + recursive(stk);
}
</details>
# 答案4
**得分**: 0
有很多种方法可以做到这一点。但是，如果您不想销毁堆栈，可以按照以下方式进行。堆栈在返回过程中恢复。
- `n` 被弹出。这使数字堆栈减少，并将 `n` 保留在本地调用堆栈中以供以后使用。
- 元素的平方保存在 `k` 中的方法调用堆栈中。
- `r` 被初始化为最终总数。
- 一旦堆栈为空，只需将 `n` 重新推回 `stk` 并返回保存的平方和。
```java
static int sum_sqr_rec(Stack<Integer> stk) {
int n = stk.pop();
int k = n*n;
int r = 0;
if (!stk.isEmpty()) {
r = sum_sqr_rec(stk);
}
stk.push(n);
return r + k;
}
使用以下语句的调用序列。
System.out.println(stack);
System.out.println(sum_sqr_rec(stack));
System.out.println(stack);
结果如下
[2, 3, 4, 5]
54
[2, 3, 4, 5]
英文:
Lots of ways to do this. But if you don't want to destroy the stack you can do it this way. The stack is restored during the return process.
-
nis popped. This depletes the stack of numbers and keepsnin the local call stack for later use. -
The square of the element is saved on the call stack of the method in
k -
ris initialized for the final tally. -
Once the stack is empty, simply push
nback onstkand return the sums of the saved squares.
static int sum_sqr_rec(Stack<Integer> stk) {
int n = stk.pop();
int k = n*n;
int r = 0;
if (!stk.isEmpty()) {
r = sum_sqr_rec(stk);
}
stk.push(n);
return r + k;
}
Using this call sequence of statements.
System.out.println(stack);
System.out.println(sum_sqr_rec(stack));
System.out.println(stack);
Results in the following
[2, 3, 4, 5]
54
[2, 3, 4, 5]
</details>
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