Question

public static void main(String[] args)
{
    Scanner scan = new Scanner(System.in);
    int n = scan.nextInt();

    for(int i = n; i >= 0; i = n / 10)
    {   
        i = i % 10;

        if(i == 1)
            System.out.print("Has 1");

        n /= 10;
    }

    System.out.print("Has No 1");
}

它正在使用＆＃34;有1＆＃34;但是当没有数字1时它没有打印任何东西。缺少什么？

Answer 1

您当前代码的逻辑将无法实现您希望实现的目标。因为它进入无限循环（对于具有1而非1个数字的两种情况），因为当mapply(function(tableau.m, filename){ p <- ggplot(tableau.m, aes(variable,Name)) + geom_tile(aes(fill = value), colour = "white") + scale_fill_distiller(palette = "YlGnBu",limits=c(min(tableau.m$value), max(tableau.m$value))) + geom_text(aes(label=value), family="AkkuratLightPro-Regular", color = "black",lineheight=.5,size = 4) base_size <- 9 p + theme_grey(base_size = base_size) + labs(x = "", y = "") + scale_x_discrete(expand = c(0, 0)) + scale_y_discrete(expand = c(0, 0)) + theme(legend.position = "none", axis.ticks = element_blank(), axis.text.x = element_text(size = 12, angle = 270, hjust = 0, colour = "grey50", family="AkkuratPro-Regular") ,axis.text.y = element_text(size = 12, angle = 0, hjust = 1, colour = "grey50", family="AkkuratPro-Regular")) + ggtitle(filename) + theme(plot.title = element_text(size = 16, angle = 0, colour = "grey25", family="AkkuratPro-Regular")) ggsave(file=paste0(filename,".png")) }, dat, matrix_add_cats_files)和i >= 0对循环中的任何正数不会为负时，您将继续循环。由于1个案例在循环中，因此它会打印该语句。

此外，即使你纠正了循环，它仍然会执行

无论是否有一个给定数字。

您需要将此代码更新为以下

System.out.print("Has No 1");

希望这有帮助

Answer 2

辅助方法使其变得简单且可重复使用：

private static String hasDigit(int n, int digit) {
    for (int i = n; i > 0; i /= 10)
        if (i % 10 == digit)
            return "Has " + digit;
    return "Has No " + digit;
}

测试

System.out.println(hasDigit(13579, 1));
System.out.println(hasDigit(24680, 1));
System.out.println(hasDigit(13579, 2));
System.out.println(hasDigit(24680, 2));

输出

Has 1
Has No 1
Has No 2
Has 2

给定正数int，如果包含1位数则打印“Has 1”，否则打印“Has No 1”

2 个答案: