使用递归向上和向下计数 [英] Counting numbers up and down using Recursion

问题描述

给出两个数字，比如说 start = 1 和 end = 4 ，我试图按顺序依次计算所有数字.不允许循环

Given two numbers, let's say start = 1 and end = 4, I am trying to count all the numbers in sequence up and then down . No looping is allowed

1 2 3 4 3 2 1

我尝试编写递归函数.该函数正在向上计数并打印1 2 3 4，但是当我尝试向下计数时，我期望 4 3 2 1 但我陷入无限循环.原因是起始值在递归中丢失，并且我不知道从下往上数时在哪里停止.

I tried writing a recursion function. The function is counting up fine and its printing 1 2 3 4 but when I try to count down, I expect 4 3 2 1 but I get into an infinite loop. The reason is the start value is lost in recursion and I don't know where to stop when counting from bottom up.

我已经花了4个小时了.我们甚至可以递归进行此操作吗?递归是一种方式

I have spent 4 hours on this . Can we even do this in recursion ? Is recursion one way

public static void countUpDown(int start, int end) {
    //to pring bottom up -> 4 3 2 1
    if ( start > end  && end > 0) {
        System.out.println(end - 1);
        countUpDown(start, end - 1);    
    }

   //to print up 1 2 3 4 
    if (start <= end) {
        System.out.println("-->" + start);
        countUpDown(start + 1, end);
    }
}

推荐答案

您只需要使用递归进行计数即可.然后，当函数返回时，您就走下了路.这可以通过以下方式实现:

You just have to use the recursion for counting up. Then, when the function returns, you are on your way down. This can be achieved with:

public void countUpAndDown(int start, int end) {
    System.out.println(start);
    if (end == start) return;
    countUpAndDown(start+1, end);
    System.out.println(start);
}

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