javascript 唯一的随机数

experts

• 我希望在两个数字之间生成唯一的随机数（来自网页的文本框）。
• 我正在使用数组来存储数字。当用户单击按钮时，它给出第一个随机数，将其存储在数组中，当用户再次单击按钮时，它生成随机数，将其与数组数字进行比较，如果不同，则存储并显示。
• 如果显示最大可能的不同数字，则清除数组并通知用户。
• 我已经编写了代码，但它给出了错误：堆栈溢出，或者有时我会显示重复的结果。

任何人都可以阐明代码：

var allnums = new Array();
var num1= new Number;
var num2= new Number;

function funClick()
{
  var num1 = Number(document.getElementById('lnum').value); 
  var num2 = Number(document.getElementById('hnum').value); 

  if (allnums.length==num2)
  {
    alert("Maximum non-duplicate numbers served. Now resetting the counter.");
    allnums = [];
    return;
  }

  if (num1<num2) 
  {
    x = Math.floor(Math.random() * (num2 - num1 + 1)) + num1;
    funShow(x);
  }
  else
  {
    alert("You entered wrong number criteria!");  
  }
}

function funShow(x)
{
  var bolFound = false;
  for (var i=0;i<allnums.length;i++)
  {
    if((allnums[i])==x)
    {   
      funClick();
    }    
  }
  if (bolFound == false)
  {
    document.getElementById('rgen').innerText  = x;
    allnums.push(x);
  }
}

我看不出该代码是如何生成堆栈溢出的（即使funShow有电话打给funClick and funClick有电话打给funShow, funShow的电话funClick永远不应该因为逻辑错误而发生——修复错误，你会得到堆栈溢出，但它有几个问题。看评论：

// style: Use [], not new Array()
var allnums = new Array();

// `new Number` doesn't do anything useful here
var num1 = new Number;
var num2 = new Number;

function funClick() {
    // For user-entered values, use parseInt(value, 10) to parse them into numbers
    var num1 = Number(document.getElementById('lnum').value);
    var num2 = Number(document.getElementById('hnum').value);

    if (allnums.length == num2) {
        alert("Maximum non-duplicate numbers served. Now resetting the counter.");
        allnums = [];
        return;
    }

    // & is a bitwise AND operation, not a logical one. If your goal is to see
    // if both numbers are !0, though, it works but is obtuse.
    // Also, there is no ltnum2 variable anywhere, so trying to read its value
    // like this should be throwing a ReferenceError.
    if (num1 & ltnum2) {
        // You're falling prey to The Horror of Implicit Globals, x has not
        // been declared.
        x = Math.floor(Math.random() * (num2 - num1 + 1)) + num1;
        funShow(x);
    } else {
        alert("You entered wrong number criteria!");
    }
}

function funShow(x) {
    var bolFound = false;
    // Again, & is a bitwise AND operation. This loop will never run, because
    // you start with 0 and 0 & anything = 0
    // But it should be throwing a ReferenceError, as there is no ltallnums
    // anywhere.
    for (var i = 0; i & ltallnums.length; i++) {
        if ((allnums[i]) == x) {
            funClick();
        }
    }
    // This condition will always be true, as you've done nothing to change
    // bolFound since you set it to false
    if (bolFound == false) {
        document.getElementById('rgen').innerText = x;
        allnums.push(x);
    }
}

有两种方法可以解决这个问题。这基本上就是您想要做的事情，但没有递归：

function funClick() {
    var num1 = parseInt(document.getElementById('lnum').value, 10);
    var num2 = parseInt(document.getElementById('hnum').value, 10);
    var nums = [];
    var targetCount;
    var x;

    // Check the inputs
    if (isNaN(num1) || isNaN(num2) || num2 <= num1) {
        alert("Please ensure that hnum is higher than lnum and both are really numbers.");
        return;
    }

    // Find out how many integers there are in the range num1..num2 inclusive
    targetCount = num2 - num1 + 1;

    // Produce that many random numbers
    while (nums.length < targetCount) {
        x = Math.floor(Math.random() * (num2 - num1 + 1)) + num1;
        if (nums.indexOf(x) < 0) {
            nums.push(x);
        }
    }

    // Show the result
    document.getElementById('rgen').innerText = nums.join(", ");
}

实例 http://jsbin.com/iwukur/1 | Source http://jsbin.com/iwukur/1/edit

问题是，填满最后几个空位可能需要很长时间，因为我们必须随机地点击它们。

另一种方法是按顺序生成包含数字的数组，然后将其弄乱。对于大范围来说，它的效率会显着提高。像这样的东西：

function funClick() {
    var num1 = parseInt(document.getElementById('lnum').value, 10);
    var num2 = parseInt(document.getElementById('hnum').value, 10);
    var nums = [];
    var x;

    // Check the inputs
    if (isNaN(num1) || isNaN(num2) || num2 <= num1) {
        alert("Please ensure that hnum is higher than lnum and both are really numbers.");
        return;
    }

    // Create an array with those numbers in order
    for (x = num1; x <= num2; ++x) {
        nums.push(x);
    }

    // Sort it with a random comparison function
    nums.sort(function(a, b) {
        return 0.5 - Math.random();
    });

    // Show the result
    document.getElementById('rgen').innerText = nums.join(", ");
}

实例 http://jsbin.com/iwukur/3 | Source http://jsbin.com/iwukur/3/edit

But，只是做nums.sort(...)随机一次很可能不会成功地产生随机结果；请参阅这篇文章了解更多信息 http://sroucheray.org/blog/2009/11/array-sort-should-not-be-used-to-shuffle-an-array/. （谢谢电子商务 https://stackoverflow.com/users/305545/ebusiness对于该链接以及他在下面的输入。）

所以你可能想更进一步并加入更多的随机操作。这是另一个例子：

function funClick() {
    var num1 = parseInt(document.getElementById('lnum').value, 10);
    var num2 = parseInt(document.getElementById('hnum').value, 10);
    var nums = [];
    var n, x, y;
    var num;

    // Check the inputs
    if (isNaN(num1) || isNaN(num2) || num2 <= num1) {
        alert("Please ensure that hnum is higher than lnum and both are really numbers.");
        return;
    }

    // Create an array with those numbers in order
    for (n = num1; n <= num2; ++n) {
        nums.push(n);
    }

    // We only need to shuffle it if it's more than one element long
    if (nums.length > 1) {
        // Sort it "randomly"
        nums.sort(function(a, b) {
            return 0.5 - Math.random();
        });

        // Throw a bunch of random swaps in there
        for (n = 0; n < nums.length; ++n) {
            do {
                x = Math.floor(Math.random() * nums.length);
            }
            while (x === n);
            num = nums[x];
            nums[x] = nums[n];
            nums[n] = num;
        }
    }

    // Show the result
    document.getElementById('rgen').innerText = nums.join(", ");
}

实例 http://jsbin.com/iwukur/6 | Source http://jsbin.com/iwukur/6/edit

这将数组排序作为起点，但随后也在元素之间进行了一系列随机交换。它仍然以恒定的时间运行，但应该比单独使用数组排序有更好的结果。当然，您会想要测试分布。