[LeetCode][JavaScript]Ugly Number II

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.

Hint:

1. The naive approach is to call isUgly for every number until you reach the nth one. Most numbers are not ugly. Try to focus your effort on generating only the ugly ones.
2. An ugly number must be multiplied by either 2, 3, or 5 from a smaller ugly number.
3. The key is how to maintain the order of the ugly numbers. Try a similar approach of merging from three sorted lists: L1, L2, and L3.
4. Assume you have Uk, the kth ugly number. Then Uk+1 must be Min(L1 * 2, L2 * 3, L3 * 5).

https://leetcode.com/problems/ugly-number-ii/

 

 

---

 

 

找出第n个ugly number，想通了不难。

动态规划，ugly number一定是由一个较小的ugly number乘以2，3或5得到的。

先开一个数组记录所有的ugly number。

然后开三个变量L1, L2, L3，一开始都是零，指向数组的第0位。

每一次取Min(L1 * 2, L2 * 3, L3 * 5)，找到这个数之后，对应的Ln加一，指向数组的下一位。

/**
 * @param {number} n
 * @return {number}
 */
var nthUglyNumber = function(n) {
    if(n === 1){
        return 1;
    }
    var arr = [1];
    var two = 0, three = 0, five = 0;
    for(var i = 1; i < n; i++){
        var min = Math.min(arr[two] * 2, arr[three] * 3, arr[five] * 5);
        arr[i] = min;
        if(arr[two] * 2 === min) two++;
        if(arr[three] * 3 === min) three++;
        if(arr[five] * 5 === min) five++;
    }
    return arr[i - 1];
};