javascript - 位串中的交替整数

Question

我正在解决一个 leetcode 问题，该问题需要编写一个公式来确定给定数字 n 的位字符串是否具有交替整数。

我觉得我的逻辑是可靠的，但我认为我定义位字符串并循环它的方式有问题。有人可以帮我找出为什么这段代码不起作用吗？谢谢，我是自学成才，这是我第一次尝试 javascript leetcode，所以任何建议都会受到赞赏!

//first we need to create a variable that is a string of the bits for whatever number is given

//create variables current and previous and set them to null
//create for loop that loops through each digit in the bit, and sets "current" the number being looped over
//is that value = to previous (null on first loop)? no, continue loop and set previous to current
//run through loop again, changing current value
//is current = previous? if yes, result = false
//if no, continue loop


//define function
var hasAlternatingBits = function(n) {
  let current = null; //set current to null so first loop iteration is always ture
  let previous = null; //previous is null so it can be changed during loop
  let result = true; //result is true until loop finds consecutive integers
  var bitString = n.toString(2).split(', '); //turn string into an array that can be looped

  //create for loop that loops entire string, or until it finds two consecutive integers
  for (let i = 0; i < bitString.length; i++) {
    //set value of current to number being looped over in string
    current = i;
    //if current doesnt equal previous, 
    if (current !== previous) {
      previous = current; //set previous to current
    } else //if current does equal previous
      result = false; //change return to false
    break; //end loop
  }

  return result;
};

console.log(hasAlternatingBits(5)) // should return true
console.log(hasAlternatingBits(7)) // should return false

Answer 1

主要问题是您将 current 分配给 i，因此它的值从 1 到 n 的长度

另请注意，您可以在循环数组时循环遍历字符串，就像我在下面的代码片段中所做的那样

---

另请注意，您的中断位于代码的 else 部分之外。

在您的代码中，如果previous === current，您可以return false，然后节省一些时间

//define function
var hasAlternatingBits = function(n) {
    let current = null; //set current to null so first loop iteration is always ture
    let previous = null; //previous is null so it can be changed during loop
    let result = true; //result is true until loop finds consecutive integers
    let bitString = n.toString(2);//turn into string
    
    //create for loop that loops entire string, or until it finds two consecutive integers
    for (let i = 0; i < bitString.length; i++) {
        //set value of current to number being looped over in string
         current = bitString[i]; // <-- here
        //if current doesnt equal previous, 
        if (current !== previous) {
            previous = current;//set previous to current
        }
        else { //if current does equal previous
            return false
        }
    }
    
    return result;
};

console.log(hasAlternatingBits(5))
console.log(hasAlternatingBits(7))