递归求解的两道小练习

 

1.求1 + 2 + 3 + ... + 100 的结果

public class TestCacu {

    public static int add(int first, int last, int result) {

        result += last;

        if (last == first) {
            return result;
        } else {
            return add(first, last - 1, result);
        }

    }
    
    public static void main(String[] args) {
        
        int res = add(0, 100, 0);
        System.out.println(res);
    }
}

 

2:一个人买汽水，一块钱一瓶汽水，三个瓶盖可以换一瓶汽水，2个空瓶可以换一瓶汽水，问20块钱总共可以买多少瓶汽水，写程序算出来

 

public class TestBuyWater {

    public static int returnBuy(int waterNum, int coverNum, int bottleNum) {

        int newWater = coverNum / 3 + bottleNum / 2;
        waterNum += newWater;
        coverNum = coverNum % 3 + newWater;
        bottleNum = bottleNum % 2 + newWater;

        if (coverNum < 3 && bottleNum < 2) {
            return waterNum;
        } else {
            return returnBuy(waterNum, coverNum, bottleNum);
        }
    }

    public static void main(String[] args) {

        int res = returnBuy(20,20,20);
        System.out.println(res);
    }
}

今天的一些练手题，涉及到了一点递归的思想。记录一下