Question

所以我有一种情况，我有两个数据，我试图从同一个for循环得到（我希望它来自相同的for循环，没有重复的代码）。

我正在寻找finalFloor，我的数据数据也会带给我;但我也在寻找地址[]中的哪个索引变量currentFloor变为负值。

下面是我的代码，目前我正在运行它作为两个独立的函数（floorCalculator＆amp; inTheBasement）运行相同的代码（不想要，而不是良好的编码实践），除了找到的最终目标。我真的在努力弄清楚如何结合这一点。任何想法或指针？谢谢你的帮助！

/* ----------------- Declaration of Variables ---------------- */
var up = '('; // represents moving up 1 floor.
var down = ')'; // represents moving down 1 floor.
var input_form = $('#input-form');  // represents the html input form.
var userInput = input_form.find('#address-input'); // represents the finding of the user's input.
var input; // stores user input value.
var address = []; // stores user's input value as an array of characters.
var currentFloor = 0; // represents the current floor in the for loop, set to ground floor (0).
var finalFloor; // represents the ending floor from the instructions given.
var results = $('.results');  // represents the div .results for appending data to.

/* ----------------- Parent Function ---------------- */
$(document).ready(initLoad);

/* ----------------- Child Functions ---------------- */
function initLoad()
{
  input_form.submit(function(event) // Listens for submission event at #input-form.
{
event.preventDefault();   // Prevents default method of html element.
takeInAddress();          // Calls function.
});
};


function takeInAddress()
{
  input = userInput.val();    // Stores the user input found at #address-input as var input.
  userInput.val('');          // Clears the input field for next user input.
  address = input.split('');  // Splits the string input into single characters stored now in the array address[ ].
  floorCalculator();          // Calls funciton.
};

function floorCalculator()
{
  for (var i = 0; i < address.length; i++)
  {
    if (address[i] == up)  // For any '(' present at the current index...
    {
      currentFloor++; // Increase the value of currentFloor by 1.
    }
    else if (address[i] == down) // For any ')' present at the current index...
    {
     currentFloor--; // Decrease the value of currentFloor by 1.
    }
  } // end for loop
  finalFloor = currentFloor; // Store the value of currentFloor now as finalFloor.
  // console.log(finalFloor);
  results.append('<h2>Floor to deliver to: ' + finalFloor + '</h2>'); // Append finalFloor value to .results html.
  inTheBasement();  // Calls function.
};

function inTheBasement()
{
  currentFloor = 0; // Resets currentFloor to zero.
  for (var i = 0; i < address.length; i++)
  {
    if (address[i] == up) // For any '(' present at the current index...
    {
      currentFloor++; // Increase the value of currentFloor by 1.
    }
    else if (address[i] == down)  // For any ')' present at the current index...
    {
      currentFloor--; // Decrease the value of currentFloor by 1.
      if (currentFloor < 0) // if currentFloor becomes a negative value...
      {
        // console.log(i);
        // Append value of i
         results.append('<h2>When you will arrive in the basement: ' + i + 'th instruction. </h2>');
         break; // break from loop
       } // end if loop
     } // end else if loop
   } // end for loop
  };

Answer 1

在第一个currentFloor < 0循环中检查for。要防止两次打印消息，请使用变量记住是否已经执行过该操作。

function floorCalculator()
{
  var foundBasement = false;
  var basementStep;
  for (var i = 0; i < address.length; i++)
  {
    if (address[i] == up)  // For any '(' present at the current index...
    {
      currentFloor++; // Increase the value of currentFloor by 1.
    }
    else if (address[i] == down) // For any ')' present at the current index...
    {
     currentFloor--; // Decrease the value of currentFloor by 1.
     if (currentFloor < 0 && !foundBasement) {
        foundBasement = true;
        basementStep = i;
     }
    }
  } // end for loop
  finalFloor = currentFloor; // Store the value of currentFloor now as finalFloor.
  // console.log(finalFloor);
  results.append('<h2>Floor to deliver to: ' + finalFloor + '</h2>'); // Append finalFloor value to .results html.
  if (foundBasement) {
    results.append('<h2>When you will arrive in the basement: ' + basementStep + 'th instruction. </h2>');
  }
};

Answer 2

因此，您的第一个循环是“reduce”的经典用例：它将数组转换为单个值。

reduce接受一个函数和一个可选的基值：

[1,2,1,1].reduce(aFunction, startValue)

我们将编写一个函数，当传递给reduce时，将一个数组的所有值加在一起。我们传递给reduce的函数应该接受两个值 - 一个'memo'，它将在函数调用之间存储状态，并在它们之间传递，一个'value'将代表数组中的下一个值，逐个传递。它应该在将值考虑之后返回状态，并且无论它返回什么，它将在下一次调用时再次传递给函数，以及数组中的下一个值。

function aFunction(value, memo) {
   return value + memo;
}
startValue = 0; // we start with 0 for our use case

我们可以缩短函数语法，如下所示：

(memo, value) => value + memo
// the return statement is implicit in this syntax

结果，传递我们的函数和我们的起始值变成了一个单一的内容：

[1,2,1,1].reduce((memo, value) => value + memo, 0)

唯一需要的其他知识是三元：

(memo, value) => value === ")" ? memo + 1 : memo - 1

以上相当于：

function (memo, value) {
    if (value === ")") {
        return memo + 1;
    }
    else {
        return memo - 1;
    }
}

最后，如果我们想要在一个reduce调用中执行此操作，我们只需要在备忘录中传递更多状态，并进行另一次评估。

ourInput = ")()()((())))))()()()(".split("");
// it's now an array, as you know
state = { floor: 0, basementTrigger: false, becameNegative: undefined };
result = ourInput.reduce( (memo, value, index) => {
    memo.floor += value === "(" ? 1 : -1; // add either 1 or negative one to our floor
    if (!memo.basementTrigger && memo.floor < 0) {
        memo.becameNegative = index
        memo.basementTrigger = true;
    }
    return memo;
}, state)  // state is passed in as 'memo' on the inner functions's first call

对于每个值，这个：

根据value是否为"("，在地板上添加或减少。
如果触发器为假，且楼层为负，则：
- 将触发器翻转为true，并存储当前索引

然后我们添加：

output += ("result = " + result.floor);
if (result.basementTrigger) output += ("follow instruction: " + result.becameNegative)

希望这有助于解决问题。

免责声明：没有校对或测试代码，可能会出错;我的目标不是为你提供代码，而是为了向你展示概念。这是一个快速抛出的hacky渲染，但应该说明你可以自己使用的工具。

在同一个for循环中求解两个变量

2 个答案: