javascript唯一随机数

我看不出该代码是如何生成堆栈溢出的(即使funShow对funClick有调用，funClick对funShow有调用，但由于逻辑错误，funShow对funClick的调用应该永远不会发生——不过，修复错误会导致堆栈溢出(，但它有几个问题。查看评论：

// 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。这里有一个基本上是你想要做的，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|源

问题是，填补最后几个空位可能需要很长时间，因为我们必须随机命中。

另一种方法是按顺序生成带有数字的数组，然后将其弄乱。对于大范围来说，它可以显著提高效率。类似这样的东西：

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|源

但是，仅仅随机进行一次nums.sort(...)很可能不会成功地产生随机结果；请参阅本文了解更多信息(感谢eBusiness的链接和他在下面的内容。(

因此，您可能想要更进一步，并投入更多的随机操作。下面是另一个例子：

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|源

这将数组排序作为一个起点，但随后还会在元素之间进行一系列随机交换。它仍然在恒定的时间内运行，但应该比单独使用数组排序有更好的结果。当然，您需要测试分布。

使用数组：

var uniqueRandomNumbers = new Array();
var totalNumbers = 100;
for (var i=0; i<totalNumbers; i++){
    uniqueRandomNumbers.push(i);
}
uniqueRandomNumbers.sort(function() {return 0.5 - Math.random();});
var uniqueNumber;
for(var i=0; i<uniqueRandomNumbers.length; i++){
    uniqueNumber = uniqueRandomNumbers[i];
    //do something with the number
}

由于我无法编辑Crowder的答案，下面是对数组进行加扰的简单无偏方法：

function scramble(nums){
    for (var n = nums.length; n; n--) {
        var x = Math.floor(Math.random() * n);
        var num = nums[n-1];
        nums[n-1] = nums[x];
        nums[x] = num;
    }
}