需要帮助来理解递归 [英] need help to understand recursion

问题描述

n = 10
10+9+8+7+6+5+4+3+2+1 = 55

这是一段代码，用于将数字从 n 开始添加到它之前的每个数字.

here's a piece of code to add number starting from n, to every number before it.

public static int recursion(int index){
    if (index == 0){
        return 0;
    }
    else{
        return recursion(index-1) + index;
    }
}

抱歉问了一个愚蠢的问题，但让我感到困惑的是:当索引不为零时，它再次调用递归函数，将索引减去 1，依此类推，直到索引为零.但是，它是编码递归(index-1) + index.

sorry for stupid question, but here's what confuse me: when the index is not zero, it calls the recursion function again, with the index substracted by 1, so on until the index is zero. However, it's coded recursion(index-1) + index.

那么为什么每次调用该函数时索引不减去1并加上10(或任何索引号)?为什么不是这样的:(10+ (9+10) + (8+10) + (7+10) +....) ?

so why the index is not substracted by 1 and added by 10 (or any index number) each time the function is called? why is it not something like this: (10+ (9+10) + (8+10) + (7+10) +....) ?

推荐答案

试着把这个总和写成

sum(10) = 1+2+3+4+5+6+7+8+9+10

这会让你看到

sum(10) = (1+2+3+4+5+6+7+8+9)+10 = sum(9)+10

和

sum(9) = (1+2+3+4+5+6+7+8)+9

等等

换句话说

sum(i) = sum(i-1) + i;

或者更准确地说

         {0            for i=0
sum(i) = {
         {sum(i-1) + i for i>0

顺便说一句，每次调用方法时，它的变量在每个方法调用中都是单独的实例，这意味着 recursion(10) 中的 index 是与 indexrecursion(9)中的code>.

BTW each time method is called its variables are separate instances in each method call which means index in recursion(10) is separate variable than index in recursion(9).

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