Question

我有一个字符串，希望在使用以下替换项时获得所有可能的replace组合：

var equiv = {
  "a": "4",
  "b": "8",
  "e": "3",
  "i": "1",
  "l": "1",
  "o": "0",
  "t": "7"
}

我想定义一个String.prototype函数，例如：

String.prototype.l33tCombonations = function()
{
    var toReturn = [];

    for (var i in equiv)
    {
        // this.???
        // toReturn.push(this???)
    }

    return toReturn;
}

所以我可以输入类似"tomato".l33tCombinations()的东西并返回：

["tomato", "t0mato", "t0mat0", "tomat0", "toma7o", "t0ma7o", "t0m470", ...].

顺序并不重要。有想法吗？

Answer 1

我将使用一种递归方法，逐个遍历字符串char：

const toL33t = { "a": "4", "b": "8",  "e": "3",  "i": "1", "l": "1",  "o": "0",  "t": "7" };

function* l33t(string, previous = "") {
  const char = string[0];
  // Base case: no chars left, yield previous combinations
  if(!char) {
    yield previous;
    return;
  }
  // Recursive case: Char does not get l33t3d
  yield* l33t(string.slice(1), previous + char);
  // Recursive case: Char gets l33t3d
  if(toL33t[char])
    yield* l33t(string.slice(1), previous + toL33t[char]);
}

console.log(...l33t("tomato"));

如果您确实还在原型possibl3上使用它，但我不建议这样做：

 String.prototype.l33t = function() {
   return [...l33t(this)];
 };

 console.log("stuff".l33t());

Answer 2

您可以使用reduce

做类似的事情

这个想法是循环遍历每个字符并将每个组合添加到累加器中。如果遇到equiv中不是部分的字符，只需将该字符添加到累加器中的每个项目中即可。如果equiv中存在字符，请复制所有先前的组合，并使用equiv[<character>]

添加另一组组合

const equiv = {
  "a": "4",
  "b": "8",
  "e": "3",
  "i": "1",
  "l": "1",
  "o": "0",
  "t": "7"
}

const input = "tomato";

const output = [...input].reduce((acc, c, i) => {
  const r = equiv[c];
  
  if (i === 0) {
    return r ? [c, r] : [c];
  }

  const updated = acc.map(a => a + c);
  const newItems = r ? acc.map(a => a + r) : [];
  
  return [...updated, ...newItems]
}, [])

console.log(output)

Answer 3

您可以采用笛卡尔积来生成所需值。

function leet(string) {
    const
        cartesian = (a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []),
        code = { a: "4", b: "8", e: "3", i: "1", l: "1", o: "0", t: "7" };

    return Array
         .from(string, c => c in code ? [c, code[c]] : [c])
         .reduce(cartesian)
         .map(a => a.join(''));
}

console.log(leet('tomatoe'));

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Answer 4

我已经以递归的方式解决了这个问题，在每次递归的迭代中，都会分析字符串的一个新字符，如果该字符有一个替换，那么character和replacement被连接到所有先前的结果，从而产生一组新的结果，否则，仅将character连接到所有先前的结果。请注意，我滥用了这种传播工具。

var equiv = {a: "4", b: "8", e: "3", i: "1", l: "1", o: "0", t: "7"};

const genComb = (str, arr) =>
{
    if (!str) return arr; // Finish condition.
    let c = str[0];       // New char to be appended.
    let r = equiv[c];     // New char replacement.

    return genComb(
        str.slice(1),
        [...arr.map(e => e + c), ...(r ? arr.map(e => e + r) : [])]
    );
};

String.prototype.l33tCombinations = function()
{
   return genComb(this, [""], 0);
}

console.log("tomato".l33tCombinations());

Answer 5

我认为这可能会产生预期的结果！遍历每个字母，每次找到新的替换字母时，都向toReturn添加一个新单词，并确保搜索每个新单词！

var equiv = {
  "a": "4",
  "b": "8",
  "e": "3",
  "i": "1",
  "l": "1",
  "o": "0",
  "t": "7"
}

String.prototype.l33tCombinations = function() {
  var toReturn = [this.toLowerCase()];

  for (let i = 0; i < toReturn.length; i++) {
  
    for (let j = 0; j < toReturn[i].length; j++) {
      if (equiv[toReturn[i][j]]) {
       let newWord = toReturn[i].split('');
       newWord[j] = equiv[newWord[j]];
       let newWordJoined = newWord.join('');
       if (!toReturn.includes(newWordJoined))
        toReturn.push(newWordJoined);
      }
    }
 
  }

  return toReturn;
}

console.log('tomato'.l33tCombinations());

用Javascript获取所有可能的l33t组合的数组

5 个答案: