LeetCode 264. Ugly Number II（丑数字）

原题网址：https://leetcode.com/problems/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.

思路：动态规划。因数2、3、5可以分别乘以前面的数字，找出最小的一个作为下一个。

public class Solution {
    public int nthUglyNumber(int n) {
        if (n<=1) return 1;
        int[] nums = new int[n];
        nums[0] = 1;
        int f2 = 0, f3 = 0, f5 = 0;
        for(int i=1; i<n; i++) {
            int m2 = nums[f2] * 2;
            int m3 = nums[f3] * 3;
            int m5 = nums[f5] * 5;
            if (m2 <= m3 && m2 <= m5) {
                nums[i] = m2;
                f2 ++;
            } else if (m3 <= m2 && m3 <= m5) {
                nums[i] = m3;
                f3 ++;
            } else {
                nums[i] = m5;
                f5 ++;
            }
            if (nums[i] == nums[i-1]) i --;
        }
        return nums[n-1];
    }
}

可以进一步优化得更简洁：

public class Solution {
    public int nthUglyNumber(int n) {
        if (n <= 1) return 1;
        int[] nums = new int[n];
        int size = 0;
        nums[size++] = 1;
        int p2 = 0, p3 = 0, p5 = 0;
        for(int i=2; i<=n; i++) {
            int m2 = nums[p2] * 2;
            int m3 = nums[p3] * 3;
            int m5 = nums[p5] * 5;
            int m = m2;
            if (m3 < m) m = m3;
            if (m5 < m) m = m5;
            nums[size++] = m;
            if (m2 == m) p2 ++;
            if (m3 == m) p3 ++;
            if (m5 == m) p5 ++;
        }
        return nums[size-1];
    }
}

另一种实现：

public class Solution {
    public int nthUglyNumber(int n) {
        if (n<=0) return 0;
        int[] factor = {2, 3, 5};
        int[] pos = new int[3];
        int[] nums = new int[n];
        nums[0] = 1;
        for(int i=1; i<n; i++) {
            int min = 0;
            if (nums[pos[1]] * factor[1] < nums[pos[min]] * factor[min]) min = 1;
            if (nums[pos[2]] * factor[2] < nums[pos[min]] * factor[min]) min = 2;
            nums[i] = factor[min] * nums[pos[min]++];
            if (nums[i] == nums[i-1]) i--;
        }
        return nums[n-1];
    }
}