Question

我必须制作乘法表，对于n = 3看起来像这样：

1 x 1 = 1 | 1 x 2 = 2 | 1 x 3 = 3 
2 x 1 = 2 | 2 x 2 = 4 | 2 x 3 = 6 
3 x 1 = 3 | 3 x 2 = 6 | 3 x 3 = 9

现在我的代码看起来像这样：

var n = 3;
var result;

for (i = 1; i <= n; i++) {
    var result = '';
    for(j = 1; j <= n; j++) {
        result += ` ${i} * ${j} = ${i * j}`;
    }
    console.log(result);
}

我的结果是：

1 x 1 = 1  1 x 2 = 2  1 x 3 = 3 
2 x 1 = 2  2 x 2 = 4  2 x 3 = 6 
3 x 1 = 3  3 x 2 = 6  3 x 3 = 9

我不知道如何添加＆＃34; |＆＃34;仅在数学运算之间分开。如果我添加＆＃34; |＆＃34;在结果变量的末尾，我也会在最后一次操作后得到它，但我不想要它。

Answer 1

你可以追加|当它不是最后一行时到最后。

var n = 3;
var result;

for (i = 1; i <= n; i++) {
    var result = '';
    for(j = 1; j <= n; j++) {
        result += ` ${i} * ${j} = ${i * j}`;
        if(j != n)
        {
          result += ' |';
        }
    }
    console.log(result);
}

Answer 2

您可以使用模运算符%或j != n添加检查，以查看它是行的最后一部分，并根据该条件添加|。

var n = 3;
var result;

for (i = 1; i <= n; i++) {
  var result = '';
  for (j = 1; j <= n; j++) {
    result += ` ${i} * ${j} = ${i * j}`;
    result += j % n ? ' |' : ''
  }
  console.log(result);
}

您可以使用Array.from()方法和join。

执行此操作

var n = 4;

const result = Array.from(Array(n), (e, i) => {
  i += 1
  return Array.from(Array(n), (a, j) => {
    j += 1
    return `${i} * ${j} = ${i * j}` + (j != n ? ' | ' : '')
  }).join('')
}).join('\n')

console.log(result)

Answer 3

您可以使用logical AND &&和字符串添加支票。它返回空字符串或分隔符。

var n = 3,
    result,
    i, j;

for (i = 1; i <= n; i++) {
    result = '';
    for (j = 1; j <= n; j++) {
        result += result && ' |';
        result += ` ${i} * ${j} = ${i * j}`;
    }
    console.log(result);
}

Answer 4

不是将管道添加到每个项目的末尾，而是将其添加到除第一个项目之外的每个项目的开头更简单：

＆＃13;

mgsub <- function(pattern, replacement, x, ...) {
  if (length(pattern)!=length(replacement)) {
    stop("pattern and replacement do not have the same length.")
  }
  result <- x
  for (i in 1:length(pattern)) {
    result <- gsub(pattern[i], replacement[i], result, ...)
  }
  result
}

isIsin <- function (identifier) {

  correctPrefix <- substr(identifier, 1, 2) %in% c(iso3166alpha2$Code, "XS")

  correctLength <- nchar(identifier) == 12  

  correctCharset <- !grepl('[[:punct:]]', identifier)

  if(!correctPrefix | !correctLength | !correctCharset) {
    return(FALSE)
  }

  # replace all character with its equivalent number  
  identifierOnlyNumbers <- mgsub(LETTERS, seq(10, 35), substr(identifier, 1, 11))

  # split the identifier in single digits and reverse its order
  characterVector <- rev(unlist(strsplit(identifierOnlyNumbers, "")))

  # Double every second digit of the group of digits with the rightmost character
  characterVector[seq(1, nchar(identifierOnlyNumbers), 2)] <- 
    as.character(as.numeric(characterVector[seq(1, nchar(identifierOnlyNumbers), 2)]) * 2)

  # Subtract 9 if > 9 (can apply to all since no digit can be greater than 9 before doubling)
  # Add up the digits
  summation <- sum(ifelse(as.numeric(characterVector) > 9, as.numeric(characterVector) - 9, as.numeric(characterVector)))

  # Take the 10s modulus of the sum, subtract it from 10 and take the 10s modulus of the result 
  # this final step is important in the instance where the modulus of the sum is 0, as the resulting check digit would be 10
  correctCheckDigit <- (10 - (summation %% 10)) %% 10 == as.numeric(substr(identifier, 12, 12))

  correctCheckDigit 

}

＆＃13;

Answer 5

我可能会使用一些range函数（例如range(3, 7) //=> [3, 4, 5, 6, 7]）然后map执行外部和内部范围，如下所示：

const range = (b, e) => Array.from({length: e - b + 1}, (_, i) => b + i)

const multTable = (m, n) => range(1, n).map(i => range(1, m).map(
  j => `${i} x ${j} = ${i * j}`
).join(' | ')).join('\n')

console.log(multTable(3, 3))

一旦您的产品或因素达到两位数，这将停止排队。如果这是一个问题，您可以将join(' | ')替换为join('\t|\t')。它不会是完美的，但它可能会更好。

Answer 6

试试这个

var n = 3;
var result;

for (i = 1; i <= n; i++) {
   var result = '';
   for(j = 1; j <= n; j++) {
      if(j!=n)
         result += ` ${i} * ${j} = ${i * j}` + '|';
      else
         result += ` ${i} * ${j} = ${i * j}`;
   }

console.log(result);
}

Answer 7

应该是：

var n = 3;
var result = '';

for (i = 1; i <= n; i++) {
    var result = '';
    for(j = 1; j <= n; j++) {
        result.length && result += '|';
        result += ` ${i} * ${j} = ${i * j}`;
    }
    console.log(result);
}

Answer 8

对于所有n：

＆＃13;

//google.com/?q=foo

＆＃13;

Answer 9

只需使用数组收集结果，然后join部分|

＆＃13;

var n = 3;
var result;

for (let i = 1; i <= n; i++) {
    result = [];
    for (let j = 1; j <= n; j++) {
        result.push(`${i} * ${j} = ${i * j}`);
    }
    console.log(result.join(' | '));
}

＆＃13;

Answer 10

另一种方法是删除“|”的最后一次出现在slice()：

的循环中

var n = 3;
var result;

for (var i = 1; i <= n; i++) {
  var result = '';
  for (var j = 1; j <= n; j++) {
    result += ` ${i} * ${j} = ${i * j} |`;
  }
  result = result.slice(0, -1);
  console.log(result)
}

Answer 11

尝试

let n =3;
[...Array(n)].map((_,i,a)=>
  console.log(a.map((_,j)=>`${i+1} * ${++j} = ${(i+1)*j}`).join(' | ')));

let n =3;
[...Array(n)].map((_,i,a)=>console.log(a.map((_,j)=>`${i+1} * ${++j} = ${(i+1)*j}`).join(' | ')));

仅在操作之间分隔

11 个答案: