[LeetCode] 264. 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. 

Example:

Input: n = 10
Output: 12
Explanation: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 is the sequence of the first 10 ugly numbers.

Note:  

1. 1 is typically treated as an ugly number.
2. n does not exceed 1690.

丑数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。首先第一个丑数是1，然后会用到三个pointer，分别代表2的倍数，3的倍数和5的倍数。1之后的每个丑数，是当前丑数分别乘以2,3和5的最小值。

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {number} n
 * @return {number}
 */
var nthUglyNumber = function(n) {
    let nums = new Array(n);
    let index2 = 0,
        index3 = 0,
        index5 = 0;
    nums[0] = 1;
    for (let i = 1; i < n; i++) {
        nums[i] = Math.min(nums[index2] * 2, Math.min(nums[index3] * 3, nums[index5] * 5));
        if (nums[i] === nums[index2] * 2) index2++;
        if (nums[i] === nums[index3] * 3) index3++;
        if (nums[i] === nums[index5] * 5) index5++;
    }
    return nums[n - 1];
};

Java实现

class Solution {
    public int nthUglyNumber(int n) {
        int[] nums = new int[n];
        int index2 = 0, index3 = 0, index5 = 0;
        nums[0] = 1;
        for (int i = 1; i < nums.length; i++) {
            nums[i] = Math.min(nums[index2] * 2, Math.min(nums[index3] * 3, nums[index5] * 5));
            if (nums[i] == nums[index2] * 2)
                index2++;
            if (nums[i] == nums[index3] * 3)
                index3++;
            if (nums[i] == nums[index5] * 5)
                index5++;
        }
        return nums[n - 1];
    }
}

LeetCode 题目总结