Java中的括号算法澄清

Question

我有以下方法生成括号的所有有效打开和关闭组合：

public static void getParentheses(int openParenthesis, int closeParenthesis, String string) {
    if (openParenthesis == 0 && closeParenthesis == 0) {
        System.out.println(string);
    }

    if (openParenthesis > closeParenthesis) {
        return;
    }

    if (openParenthesis > 0) {
        System.out.println("openParenthesis: " + openParenthesis + " / closeParenthesis: " + closeParenthesis);
        getParentheses(openParenthesis - 1, closeParenthesis, string + "(");
    }

    if (closeParenthesis > 0) {
        System.out.println("openParenthesis: " + openParenthesis + " / closeParenthesis: " + closeParenthesis);
        getParentheses(openParenthesis, closeParenthesis - 1, string + ")");
    }
}

我用这一行称这个方法：

getParentheses(3, 3, "");

输出是：

openParenthesis: 3 / closeParenthesis: 3
openParenthesis: 2 / closeParenthesis: 3
openParenthesis: 1 / closeParenthesis: 3
openParenthesis: 0 / closeParenthesis: 3
openParenthesis: 0 / closeParenthesis: 2
openParenthesis: 0 / closeParenthesis: 1
((()))
openParenthesis: 1 / closeParenthesis: 3 //Line8
openParenthesis: 1 / closeParenthesis: 2
openParenthesis: 0 / closeParenthesis: 2
openParenthesis: 0 / closeParenthesis: 1
(()())

我不明白为什么在Line8，openParenthesis具有1的价值，因为该方法的最后一次调用是与0和0？是什么让它增加到1？

Answer 1

第3行和第8行来自相同的函数调用：第3行由

if (openParenthesis > 0) {
    System.out.println("openParenthesis: " + openParenthesis + " / closeParenthesis: " + closeParenthesis);
    getParentheses(openParenthesis - 1, closeParenthesis, string + "(");
}

第8行：

if (closeParenthesis > 0) {
    System.out.println("openParenthesis: " + openParenthesis + " / closeParenthesis: " + closeParenthesis);
    getParentheses(openParenthesis, closeParenthesis - 1, string + ")");
}

从递归调用返回后。