Question

我想打印一个类似的模式：

This is Required pattern

到目前为止，我只能实现奇数这样的数字：

1 
3 5 7

-

Scanner kb = new Scanner(System.in);
int num = kb.nextInt();
while (num % 2 == 0 || num < 0) {
    num = kb.nextInt();
}
int odd = 1;
for (int i = 1; i <= num; i += 2) {
    String a = "";
    for (int j = 1; j <= i; j++) {
        a = odd + " ";
        odd += 2;
        System.out.print(a);
    }
    System.out.println();
}

我是初学者和新学员。请帮忙

Answer 1

我不确定预期的结果是什么，因为该模式不明确，但这可能是您正在寻找的：

Tags

结果：

int evenCounter = 1;
int oddCounter = 2;
for (int i = 1; i <= 10; i++) {
   boolean even = (i % 2 == 0);
   for (int j = 1; j < i; j++) {
      System.out.print((even ? evenCounter : oddCounter) + " ");
      evenCounter += even ? 2 : 0;
      oddCounter +=  even ? 0 : 2;
   }
   System.out.println();
}

如果每行的长度很重要，那么第二个for循环应该有一个不同的退出条件，我想

Answer 2

要从3 5 7获取模式，此代码可以执行，其中MAX_PATTERN_NUM是模式最后一行的第一个数字（在您的示例中为MAX_PATTERN_NUM = 15）

 for(int i = 3; i <= MAX_PATTERN_NUM ; i+=3)
        for(int j = i; j <= (i + 4); j+=2)
            System.out.print(j + " ");
        System.out.println();

但是，我看不到使用相同的嵌套for循环获取整个模式的逻辑方法，所以我希望这有帮助

Answer 3

这个怎么样：

public void printPattern(){

    int evenCounter = 2;
    int oddCounter = 1; //counters

    Scanner kb = new Scanner(System.in);
    int num = kb.nextInt();
    while (num % 2 == 0 || num < 0) {
        num = kb.nextInt(); //input
    }

    for(int i = 1; i <= num ; ++i){
        if(i % 2 == 0)
            evenCounter = addNumbers(i, evenCounter); //print line with evenCounter
        else
            oddCounter = addNumbers(i, oddCounter); //print line with oddCounter
    }
}

private int addNumbers(int i, int counter){
    for(int j = 0; j < i;){
        if(getIntLength(counter) + j > i) //if the number to long
            System.out.print(cut(counter, getIntLength(counter) + j - i) + " "); //output the cut version
        else
            System.out.print(counter + " ");
        j += getIntLength(counter);
        counter += 2;
    }
    System.out.println();
    return counter;
}

private String cut(int i, int length){
    return Integer.toString(i).substring(0,length); //get substring of int
}

private int getIntLength(int i){
    return Integer.toString(i).length(); //get the length of int value
}

这不是那么简单，但我没有看到更简单的方法

我的回答是基于假设最后2是截止21

Answer 4

int n,i,j,o=1,e=2;
System.out.println("enter n:");
n=sc.nextInt();
for(i=1;i<=n;i++){
    for(j=1;j<=i;j++){

        if(i%2!=0){
            System.out.print(o);
            o+=2;
        }

        if(i%2==0){
            System.out.print(e);
            e+=2;
        }
    }
    System.out.println();
}

Java模式1行奇数和1行偶数

4 个答案: