理解Java中的递归 [英] Understanding Recursion in Java

问题描述

我很难理解基于Java中的递归算法的以下代码。我不明白，当他们互相打电话时， x 和 y 的价值是多少？我试图通过在代码中调用 System.out.print（）来获得正确的值，但仍然没有得到帮助。

I'm having a tough time in understanding the following code based on recursion algorithm in Java. I don't understand, what's the different value x and y have when they call within one another? I tried to get the right value by calling System.out.print() within a code, but still get no help.

public class RecursionExample
{

    private static int[][] arr={
        {3},
        {7, 4},
        {2, 4, 6},
        {8 ,5, 9, 3}
    };

    public static int maxSum(int[][] graph, int x, int y, int sum) {
        if (x == 3) 
        {
            return sum+graph[x][y];
        }
        int max= Math.max(maxSum(graph, x+1, y, sum), maxSum(graph, x+1, y+1, sum));

        sum += graph[x][y];
        return sum+max;
    }

    public static void main(String[] ar)
    {
        System.out.println(maxSum(arr,0,0,0));
    }
}

我不是编程大师，我我想通过这样做来学习Java。任何帮助都表示赞赏。

I'm not a master in programming, and I'm trying to learn Java by doing. Any help is appreciated.

推荐答案

基本上，这会一直调用自己直到你到达第三次迭代（ x == 3 ）。

Essentially, this keeps calling itself until you reach the third iteration (x==3).

所以，这是流量（减去对 maxSum 在 max ...为了简单起见）（每个缩进是对 maxSum 的调用）：

So, here's the flow (minus the two calls to maxSum within max...for simplicity's sake) (each indent is a call to maxSum):

x = 0
y = 0
sum = 0

x != 3

    x = 1
    y = 0
    sum = 0

    x != 3

       x = 2
       y = 0
       sum = 0

       x != 3

           x = 3
           y = 0
           sum = 0

           x == 3
           return 0 + 8 //graph[3][0] == 8

       max = 8 //previous return
       sum = 0 + 2 //graph[2][0] == 2
       return 10 //max + sum == 8 + 2 == 10

    max = 10
    sum = 0 + 7 //graph[1][0] == 7
    return 17

max = 17
sum = 0 + 3 //graph[0][0] == 3
return 20

这篇关于理解Java中的递归的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！