Ugly Number II

Write a program to find the n-th ugly number.
Ugly numbers are positive numbers whose prime factors only include 2, 3, 5.
For example, 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.
Note that 1 is typically treated as an ugly number.
找到第n個(gè)丑陋數(shù)。
所有的丑陋數(shù)都是從1開(kāi)始乘2,3,5得到的，所以我們?cè)谝焉傻臄?shù)中找到乘以2,3,5之后比所有已生成數(shù)大的中最小的那個(gè)就是下一個(gè)數(shù)。
我們要怎么找這個(gè)數(shù)呢，把所有已生成數(shù)存在數(shù)組里，第一個(gè)數(shù)是1。
我們要找的下一個(gè)數(shù)一定是：

min(results[i] * 2, results[j] * 3, results[k] * 5)

這里的i，j，k怎么確定？
如果最后一個(gè)丑陋數(shù)是使用i*2生成的，那么下一次一定就不用找i了。所以：

var nthUglyNumber = function(n) {
    var results = [1];
    var i = 0, j = 0, k = 0;
    while (results.length < n)
    {
        results.push(Math.min(results[i] * 2, results[j] * 3, results[k] * 5));
        if (results[results.length - 1] === results[i] * 2) ++i;
        if (results[results.length - 1] === results[j] * 3) ++j;
        if (results[results.length - 1] === results[k] * 5) ++k;
    }
    return results.pop();
};

Super Ugly Number

Write a program to find the nth super ugly number.
Super ugly numbers are positive numbers whose all prime factors are in the given prime list primes of size k.
For example, [1, 2, 4, 7, 8, 13, 14, 16, 19, 26, 28, 32] is the sequence of the first 12 super ugly numbers given primes = [2, 7, 13, 19] of size 4.
Note:
(1) 1 is a super ugly number for any given primes.
(2) The given numbers in primes are in ascending order.
(3) 0 < k ≤ 100, 0 < n ≤ 106, 0 < primes[i] < 1000.

這題和上題差不多，只不過(guò)質(zhì)數(shù)不是固定的，是參數(shù)給出的，那么使用和剛才一樣的思想：

var nthSuperUglyNumber = function(n, primes) {
    var num = primes.length;
    var indexs = [];
    for (var i = 0;i<num;i++) 
        indexs.push(0);
    var result = [1];
    for (n;n>1;n--) {
        var min = Number.MAX_VALUE;
        for (var j = 0;j < num;j++) {
            if (result[indexs[j]] * primes[j] < min) {
                min = result[indexs[j]] * primes[j];
            }
        }
        for (j = 0;j < num;j++) {
            if (result[indexs[j]] * primes[j] == min) {
                indexs[j]++;
            }
        }
        result.push(min);
    }
    return result.pop();
};