[LeetCode] 264. Ugly Number II

丑陋数字II。题目问的是求出第N个丑陋数字。思路是需要按规则求出前N-1个丑陋数字才行。代码有点像DP但又不是DP，应该比较直观。

时间O(n)

空间O(n)

 1 /**
 2  * @param {number} n
 3  * @return {number}
 4  */
 5 var nthUglyNumber = function(n) {
 6     let nums = new Array(n);
 7     let index2 = 0,
 8         index3 = 0,
 9         index5 = 0;
10     nums[0] = 1;
11     for (let i = 1; i < n; i++) {
12         nums[i] = Math.min(nums[index2] * 2, Math.min(nums[index3] * 3, nums[index5] * 5));
13         if (nums[i] === nums[index2] * 2) index2++;
14         if (nums[i] === nums[index3] * 3) index3++;
15         if (nums[i] === nums[index5] * 5) index5++;
16     }
17     return nums[n - 1];
18 };

标签：nums,++,Number,II,Ugly,let,index2,index3,index5
来源： https://www.cnblogs.com/aaronliu1991/p/11717218.html