FreeCodeCamp挑戰：解釋錯誤消息？

退房我的解決方案爲FreeCodeCamp's Advanced Algorithm: No Repeats challenge：FreeCodeCamp挑戰：解釋錯誤消息？

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

正確的代碼應該返回。有人可以向我解釋這些錯誤信息嗎？

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：爲了獲得在錯誤的確切行號，複製&這裏粘貼代碼：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

來源

2016-11-08 veron

當您處於無限循環時會發生這種情況。檢查你的遞歸，以確保它應該退出。 –

tibsar指出你和我在正確的方向。通過「aa」或「ff」findFactorial將繼續調用自身，直到堆疊吹。除了檢查負輸入（檢查非整數輸入）之外，您還可以添加bulletproofing以查找Factorial。但與此同時，我發現在你的代碼有問題的行：你需要向右滾動才能看到壞的部分和需要改變什麼，你忘了。長度的兩倍]

   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]));

它應該是，

   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，這似乎並未給出正確的答案，但它確實解決了您詢問的錯誤。

來源

2016-11-08 04:09:36

謝謝，解決了！ – veron

FreeCodeCamp挑戰：解釋錯誤消息？

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