leetcode 264.丑数II Java

做题博客链接

https://blog.youkuaiyun.com/qq_43349112/article/details/108542248

题目链接

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

描述

编写一个程序，找出第 n 个丑数。

丑数就是质因数只包含 2, 3, 5 的正整数。

说明:  

1 是丑数。
n 不超过1690。

示例

输入: n = 10
输出: 12
解释: 1, 2, 3, 4, 5, 6, 8, 9, 10, 12 是前 10 个丑数。

初始代码模板

class Solution {
    public int nthUglyNumber(int n) {

    }
}

代码

推荐题解：
https://leetcode-cn.com/problems/ugly-number-ii/solution/chou-shu-ii-by-leetcode-solution-uoqd/

优先队列

class Solution {
    public int nthUglyNumber(int n) {
        PriorityQueue<Long> queue = new PriorityQueue<>();
        queue.offer(1L);
        Set<Long> set = new HashSet<>();
        set.add(1L);

        long[] tmp = new long[3];
        long cur = 0;
        for (int i = 1; i < n; i++) {
            cur = queue.poll();
            init(tmp, cur);
            for (int j = 0; j < tmp.length; j++) {
                if (!set.contains(tmp[j])) {
                    set.add(tmp[j]);
                    queue.offer(tmp[j]);   
                }
            }
        }

        cur = queue.poll();
        return (int)cur;
    }

    private void init(long[] tmp, long cur) {
        tmp[0] = cur * 2;
        tmp[1] = cur * 3;
        tmp[2] = cur * 5;
    }
}

动态规划

class Solution {
    public int nthUglyNumber(int n) {
        int[] dp = new int[n + 1];
        dp[1] = 1;

        //三指针，a对应2，b对应3，c对应5
        int a = 1;
        int b = 1;
        int c = 1;
        //三个中间变量
        int num1 = 0;
        int num2 = 0;
        int num3 = 0;
        for (int i = 2; i <= n; i++) {
            num1 = dp[a] * 2;
            num2 = dp[b] * 3;
            num3 = dp[c] * 5;

            dp[i] = Math.min(Math.min(num1, num2), num3);
            if (dp[i] == num1) {
                a++;
            }
            if (dp[i] == num2) {
                b++;
            }
            if (dp[i] == num3) {
                c++;
            }
        }

        return dp[n];
    }
}