LeetCode-263.264. Ugly Number II (JAVA)丑数

263. Ugly Number

Write a program to check whether a given number is an ugly number.

Ugly numbers are positive numbers whose prime factors only include 2, 3, 5. For example, 6, 8 are ugly while 14 is not ugly since it includes another prime factor 7.

Note that 1 is typically treated as an ugly number

判断一个数字是不是丑数

	public boolean isUgly(int num) {
		// num & 1 == 0
		if (num == 0)
			return false;
		while (num % 2 == 0)
			num = num >> 1;
		while (num % 3 == 0)
			num /= 3;
		while (num % 5 == 0)
			num /= 5;
		return num == 1;

	}

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. 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, and n does not exceed 1690.

寻找第n个丑数（丑数只有2，3，5的最小因子)

动态规划：

	public int nthUglyNumber(int n) {
		int[] dp = new int[n];
		dp[0] = 1;
		// 2, 3, 5最小丑数
		int f2 = 2, f3 = 3, f5 = 5;
		// f2, f3, f5对应最小值的下标
		int ix2 = 0, ix3 = 0, ix5 = 0;

		for (int i = 1; i < n; ++i) {
			int min = Math.min(Math.min(f2, f3), f5);
			// dp[i]设置为最小值
			dp[i] = min;
			// 下方if只能执行一个，求出下一个丑数可能为最小值，
			// 用于和其他未改变两个的值大小
			// 同时更新对应的下标
			// 注意ix2，3，5是从下标0开始
			// 若为1，则是后++
			if (min == f2)
				f2 = 2 * dp[++ix2];
			if (min == f3)
				f3 = 3 * dp[++ix3];
			if (min == f5)
				f5 = 5 * dp[++ix5];
		}

		return dp[n - 1];
	}

	public int nthUglyNumber(int n) {
		int[] dp = new int[n];
		// 第一个丑数是1
		dp[0] = 1;
		// 对应2，3，5的下标
		int idx2 = 0;
		int idx3 = 0;
		int idx5 = 0;
		int counter = 1;
		while (counter < n) {
			// 找出数组dp[idx2]*2、
			//dp[idx3]*3、dp[idx5]*5的最小值，
			// 最小值即为下一个丑数，同时更新最小值对应的下标，
			// 如果多个数字同时为最小值，则它们的下标都要更新
			int min = minOf
					(dp[idx2] * 2, 
					dp[idx3] * 3,
					dp[idx5] * 5);

			if (min == dp[idx2] * 2) {
				idx2++;
			}
			if (min == dp[idx3] * 3) {
				idx3++;
			}
			if (min == dp[idx5] * 5) {
				idx5++;
			}
			dp[counter] = min;
			counter++;
		}
		return dp[n - 1];
	}

	private int minOf(int a, int b, int c) {
		int temp = a < b ? a : b;
		return temp < c ? temp : c;
	}

	// 使用队列的动态规划
	// O(n) (might be more) time, O(3n) space
	public int nthUglyNumber(int n) {
		Queue q1 = new LinkedList<>();
		Queue q2 = new LinkedList<>();
		Queue q3 = new LinkedList<>();
		// 把第一个丑数加进去
		q1.offer(1);
		q2.offer(1);
		q3.offer(1);

		int m = 0;
		for (int i = 0; i < n; ++i) {
			m = Math.min(Math.min(q1.peek(), q2.peek()), q3.peek());
			// i=0时，队列头部元素都为1，此时三个队列同时更新最小值
			if (m == q1.peek())
				q1.poll();
			if (m == q2.peek())
				q2.poll();
			if (m == q3.peek())
				q3.poll();
			q1.offer(2 * m);
			q2.offer(3 * m);
			q3.offer(5 * m);
		}

		return m;
	}

参考：https://leetcode.com/problems/ugly-number-ii/#/solutions