在递归中使用基本情况多次获得相同的值

Question

//aim -> Given a number n, return the number of steps required to reach 1.
//ex -> when n=12, we get 9 steps


// export 
const steps = (num) => {
  //function to check if the number is positive integer
  isValid=n=>n>0;
  

  if(isValid(num)){
    //base case -> the number becomes 1
    if(num===1){
      return 1;
    }
    //if the number is not equal to 1
    else{
      //if the number is even do n/2 and return it again
      if(num%2===0){
        return steps(num/2);
      }
      else{

        return steps(3*num+1);
      }
    }
  }

};
console.log( steps(12) )
/**
 * lets do with recursion:
 * take an input num
 * if the number is 1, return 1 (base case)
 * if the number is not 1 but even, do cal=num/2  and return step(cal) back to the function
 * if the number is not 1 bu odd, do cal = 3*num+1
 */

当我控制台记录 isEven 时，我得到 1-> 3 次，当我执行 isOdd console.log 时，我得到了很多次。

这里有什么问题？以后我该如何预防？谢谢！

Answer 1

自己看：您的案例只考虑计算值，不计算步数

const cl = (n, op ,v )=> console.log(`${n} -->${op} =  ${v}`)

const steps = num =>
  {
  if(num > 0)
    {
    if (num===1) { cl(1,'-',1);  return 1 }
    else
      {
      if (num%2===0) { cl(num, 'n/2',   num/2  );  return steps(num/2)   }
      else           { cl(num, '3*n+1', 3*num+1);  return steps(3*num+1) }
      }
    }
  else return 'invalid num'
  }
  
console.log( 'steps(12) =', steps(12) )

.as-console-wrapper {max-height: 100%!important;top:0 }

一个“正确”的解决方案：

const stepsCount = (num, step=0) =>
  {
  step++
  if(num > 0)
    {
    if (num===1)  return step 
    else
      {
      if (num%2===0) return stepsCount(num/2,   step)   
      else           return stepsCount(3*num+1, step) 
      }
    }
  else return 'invalid num'
  }
 
console.log( 'stepsCount(12) =', stepsCount(12) )

Answer 2

您的日志结果是被调用函数的最后一次迭代的输出。 如果通过此函数跟踪num，您将获得所有迭代

const steps = (num) => {
 //function to check if the number is positive integer
 isValid=n=>n>0;
 console.log("e",num);

 if(isValid(num)){
 //base case -> the number becomes 1
 if(num===1){
  return 1;
 }
 //if the number is not equal to 1
 else{
  //if the number is even do n/2 and return it again
  if(num%2===0){
    return steps(num/2);
  }
  else{

    return steps(3*num+1);
  }
 }
}

};
console.log( steps(12) );