Google Code Jam 2016：如何优化代码？

Question

Problem:
Bleatrix Trotter the sheep has devised a strategy that helps her fall asleep faster.
First, she picks a number N. Then she starts naming N, 2 × N, 3 × N, and so on. Whenever she names a number, she thinks about all of the digits in that number.
She keeps track of which digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) she has seen at least once so far as part of any number she has named.
Once she has seen each of the ten digits at least once, she will fall asleep. Bleatrix must start with N and must always name (i + 1) × N directly after i × N. For example, suppose that Bleatrix picks N = 1692. She would count as follows: N = 1692. Now she has seen the digits 1, 2, 6, and 9. 2N = 3384. Now she has seen the digits 1, 2, 3, 4, 6, 8, and 9. 3N = 5076. Now she has seen all ten digits, and falls asleep.
What is the last number that she will name before falling asleep? If she will count forever, print INSOMNIA instead.

https://code.google.com/codejam/contest/6254486/dashboard

Array.prototype.unique = function () {
    return this.filter(function (value, index, self) {
        return self.indexOf(value) === index;
    });
}

var uniqueArr = [];
var Number = 1692;//Any number 


for (i = 1; ; i++) {
    var x = Number * i;
    while (x > 0) {
        uniqueArr.push(x % 10); //Converting number to Digits and pushing them into an array.
        x = Math.floor(x / 10);
    }
    var ar = uniqueArr.unique();
    if (ar.length == 10) {
        console.log(uniqueArr.unique(), Number * i);
        break;
    }
}

Answer 1

As long as x is always a positive number* (which is the case here) you could use

x = ~~(x / 10)

instead of

x = Math.floor(x / 10)

Which does the same job but is ~5 times faster. Its more an obvious improvement but it optimize the code anyway.

* ~~ is a bitwise operator, which just cuts off everything after the comma. So ~~-6.6 will result -6, but Math.floor(-6.6) gives you a -7. Be carefull here.