FreeCodeCamp挑战：解释错误消息？

退房我的解决方案为FreeCodeCamp's Advanced Algorithm: No Repeats challenge：FreeCodeCamp挑战：解释错误消息？

Return the number of total permutations of the provided string that don't have repeated consecutive letters.

正确的代码应该返回。有人可以向我解释这些错误信息吗？

RangeError: Maximum call stack size exceeded
at findFactorial:14:24
at findFactorial:21:14
at findFactorial:21:14
at findFactorial:21:14
at findFactorial:21:14
at findFactorial:21:14

注：为了获得在错误的确切行号，复制&这里粘贴代码：https://repl.it/

function permAlone(str) { 
     var final, factorial, repeated, i; 
     repeated = str.match(/([a-z])(?:.*)(\1)+/g); 
     if (str.length < 2) { 
      return 1; 
     } 

     // should return ["aa", "ff"] 
     if (repeated[0] === str) { 
      repeated[0] = repeated[0].split('').sort().join('').match(/([a-z])(?:.*)(\1)+/g); 
      repeated = repeated.reduce(function(a, b) { 
       return a.concat(b); 
      }); 
     } 

     function findFactorial(n) { 
      if (n < 0) { 
       alert("No negative numbers accepted."); 
      } 
      if (n === 0) { 
       return 1; 
      } 
      return n * findFactorial(n - 1); 
     } 

     factorial = findFactorial(str.length); // 7! = 5040 

     for (i = 0; i < repeated.length; i++) { 
      i++; 
      if (repeated.length === 1 && repeated.join("") !== str) { 
       final = factorial - findFactorial((str.length - 1)) * findFactorial(repeated[0].length); 
      } else if (repeated.length > 1 && repeated[i-1].length>2 || repeated[i].length>2) { 
       final = findFactorial(repeated[i].length) * findFactorial(repeated[i - 1].length); 
      } else { 
       final = factorial - ((findFactorial((str.length - 1) * repeated[i].length) * (findFactorial(str.length - 1) * repeated[i - 1].length))) + (findFactorial(str.length - 2) * findFactorial(repeated[i - 1]) * findFactorial(repeated[i])); 
       // final = 5040 - ((6! * 2!)*2) + (5! * 2! * 2!); 
      } 
     } 
     return final; 
    } 
    permAlone('abfdefa'); // should return 2640

来源

2016-11-08 veron

当您处于无限循环时会发生这种情况。检查你的递归，以确保它应该退出。 –

tibsar指出你和我在正确的方向。通过“aa”或“ff”findFactorial将继续调用自身，直到堆叠吹。除了检查负输入（检查非整数输入）之外，您还可以添加bulletproofing以查找Factorial。但与此同时，我发现在你的代码有问题的行：你需要向右滚动才能看到坏的部分和需要改变什么，你忘了。长度的两倍]

   final = factorial - ((findFactorial((str.length - 1) * repeated[i].length) * (findFactorial(str.length - 1) * repeated[i - 1].length))) + (findFactorial(str.length - 2) * findFactorial(repeated[i - 1]) * findFactorial(repeated[i]));

它应该是，

   final = factorial - ((findFactorial((str.length - 1) * repeated[i].length) * (findFactorial(str.length - 1) * repeated[i - 1].length))) + (findFactorial(str.length - 2) * findFactorial(repeated[i - 1].length) * findFactorial(repeated[i].length));

碰巧，这似乎并未给出正确的答案，但它确实解决了您询问的错误。

来源

2016-11-08 04:09:36

谢谢，解决了！ – veron

FreeCodeCamp挑战：解释错误消息？

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