Question

public static void main(String[] args) {
    int num;
    int large;
    int small;
    int secondLarge;

    Scanner scan = new Scanner(System.in);
    System.out.print("Input a number: ");
    num = scan.nextInt();
    large = num;
    small = num;
    secondLarge = num;

    for (int x = 9; x > 0; x--) {
        System.out.print("Enter " + x + " more number: ");
        num = scan.nextInt();

        if (num > large) {
            large = num;

        }
        if (num > secondLarge) {
            secondLarge = num;
        }

        if (secondLarge > large) {
            large = secondLarge;

        }
        if (num < small) {
            small = num;

        }

    } System.out.println( large + " is the largest number, " + secondLarge + " is the second largest number and " + small + " is the smallest number!");
}

所以，我试图输出最大数量，第二大数量和最小数量。我能够获得最小和最大，但不知道从哪里开始获得第二大。我认为这会起作用，但输出同样大的东西。我目前不想使用数组，因为这是家庭作业。请不要告诉我确切的答案，但是一些帮助和提示会很精彩！

Answer 1

初始化像

这样的变量

int largest=0;
int secondlargest=0;
int smallest=Integer.MAX_VALUE;

条件应该像

if(number>=largest){
    secondlargest=largest;
    largest=number;
}else if(number>secondlargest){
    secondlargest=number;
}
if(number<smallest){
    smallest=number;
}

Answer 2

这是我使用来自Math类的静态方法max和min的解决方案：

newbutton

Answer 3

public static void main(String args[]) {
    Scanner in = new Scanner(System.in);
    int max = 0;
    int smax = 0;
    int num = 0;
    while (num != -1) {
        System.out.println("Enter number  Or -1 to Stop:");
        num = in.nextInt();
        if (num >= max) {
            smax = max;
            max = num;
        } 
    }
    
    System.out.println("second max :"+smax);

}

如何在for循环中找到第二大数字

3 个答案: