Javascript Pangram 正则表达式

Javascript Pangram Regex

我正在尝试编写一个 REGEX 来测试 PANGRAM。我可以用传统的方式来做,但似乎无法用正则表达式解决超过 90% 的测试。

输入:字符串

输出:真||假

function isPangram(string){ 
   return ___________________.test(string) 
}

到目前为止的测试结果。

6/10 /([a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z, \s]+)/i

6/10 /[a-z]{1}/i

6/10 /[a-z]/i

6/10 /[a-z]+/i

9/10 /a?b?c?d?e?f?g?h?i?j?k?l?m?n?o?p?q?r?s?t?u?v?w?x?y?z/i 仅针对 abcdefghijklmopqrstuvwxyz

失败

6/10 /[\w.]+/

非常感谢任何帮助或建议。

  1. 将字符串转换为小写
  2. 使用正则表达式从字符串中提取所有唯一的字母表
  3. 检查唯一字母的数量是否为 26

代码:

function isPangram(string) {
    var regex = /([a-z])(?!.*)/g;
    return (string.match(regex) || []).length === 26;
}

Regex101

var regex = /([a-z])(?!.*)/g;

function check() {
  var val = document.getElementById('text').value.toLowerCase();

  alert(val.match(regex).length == 26);
}
<input type="text" id="text" />

<button onclick="check()">Check</button>

这将是挑战的正确答案:

function isPangram(string){ 
   return /(?=.*a)(?=.*b)(?=.*c)(?=.*d)(?=.*e)(?=.*f)(?=.*g)(?=.*h)(?=.*i)(?=.*j)(?=.*k)(?=.*l)(?=.*m)(?=.*n)(?=.*o)(?=.*p)(?=.*q)(?=.*r)(?=.*s)(?=.*t)(?=.*u)(?=.*v)(?=.*w)(?=.*x)(?=.*y)(?=.*z)./i.test(string) 
}

它对每个字母使用前瞻来检查它们是否在传递的字符串中。

如果您正在寻找非正则表达式解决方案

const isPangram = (string) => 
     new Set(string.toLowerCase().match(/[a-z]/g)).size === 26;

console.log(isPangram("abcdefghijklmnopqrstuvwxyz"));  //true
console.log(isPangram("The Quick Brown Fox Jumps over the lazy dog")); //true

function isPangram(string){

  let lc = string.toLowerCase()
 let alphabet = "abcdefghijklmnopqrstuvwxyz"

   return alphabet.split("").filter(c => lc.indexOf(c) === -1).length === 0;
}
  1. 小写字符串
  2. 用字母创建一个字符串
  3. 拆分字母表
  4. 将 filter() 与迷你函数一起使用,该函数将验证其中是否存在 0 个错误语句(对其进行过滤,以便过滤器方法创建的新数组中有 0(零)个错误语句 (-1)“ .lengeth === 0" 确保没有虚假陈述 )

作为单个正则表达式:

/(?:(?=(.*?([a-z]))(?!.*))){26}/i

regex101 进行测试。

细分:

/
  (?:               // non-capturing group
    (?=             // look-ahead assertion

      (.*?([a-z]))  // match a letter, preceded by as few characters as possible
      (?!.*)      // assert that this letter does not appear in the remainder of the string
                    // (in aggregate, this will find the last occurrence of each letter)

    )             // actually match that group (this is done to prevent backtracking)
  ){26}             // match 26 times (once for each letter)
/i                  // match case-insensitively

如果您使用的正则表达式引擎支持 atomic grouping(例如 PCRE),则可以更简洁地编写:

/(?>.*?([a-z])(?!.*)){26}/i

regex101.

给定一个字符串,检测它是否是 pangram。 Return如果是则为真,否则为假。忽略数字和标点符号。 [根据@Tushar提供的正则表达式添加不区分大小写]

//Detect Pangram
function isPangram(string){
// character set capturing group with negative lookahead
  let regex = /([a-z])(?!.*)/gi;
  return (string.match(regex)).length === 26;
}

console.log(isPangram("The quick brown fox jumps over the lazy dog."));// true
console.log(isPangram("This is not a pangram."));// false
console.log(isPangram("Pack my box with five dozen liquor jugs."));// true
console.log(isPangram("This isn't a pangram!"));// false
console.log(isPangram("Detect Pangram"));// false
console.log(isPangram("How quickly daft jumping zebras vex."));// true

function isPangram(input) {
  if(input.length < 26) {
    return false;
  }
  const letters = 'abcdefghijklmnopqrstuvwxyz';
  return Array.from(new Set(input.toLowerCase().split('').filter(i => i.trim()))).sort().join('') === letters;  
}

console.log(isPangram("The String is abcdefghijklumnopqrstvwxyz")); //true

使用 for..of 循环 & 包括:

function isPangram(sentence) {
let lowerCased = sentence.toLowerCase();
for (let char of 'abcdefghijklmnopqrstuvwxyz') {
    if (!lowerCased.includes(char)) {
        return false;
     }
  }
 return true;
}

const isPangram = (string) => 
     new Set(string.toLowerCase().match(/[a-z]/g)).size === 26;

console.log(isPangram("This is not a pangram."));  //false
console.log(isPangram("The Quick Brown Fox Jumps over the lazy dog")); //true