【问题描述】[中等]

【解答思路】

1. 动态规划

时间复杂度：O(N) 空间复杂度：O(N)

class Solution {public int nthUglyNumber(int n) {int a = 0, b = 0, c = 0;int[] dp = new int[n];dp[0] = 1;for(int i = 1; i < n; i++) {int n2 = dp[a] * 2, n3 = dp[b] * 3, n5 = dp[c] * 5;dp[i] = Math.min(Math.min(n2, n3), n5);if(dp[i] == n2) a++;if(dp[i] == n3) b++;if(dp[i] == n5) c++;}return dp[n - 1];}
}作者：jyd
链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/mian-shi-ti-49-chou-shu-dong-tai-gui-hua-qing-xi-t/

2. 堆

时间复杂度：O(N) 空间复杂度：O(1)

class Solution {public int nthUglyNumber(int n) {PriorityQueue<Long> pq = new PriorityQueue<>();Set<Long> s = new HashSet<>();//初始化,放进堆和set，发现1要开Long数组才可以long[] primes = new long[]{2, 3, 5};for (long prime : primes) {pq.offer(prime);s.add(prime);}long num = 1;for (int i = 1; i < n; i++) {num = pq.poll();//遍历三个因子for (int j = 0; j < 3; j++) {if (!s.contains(num * primes[j])) {pq.offer(num * primes[j]);s.add(num * primes[j]);}}}return (int) num;}
}作者：jerry_nju
链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/dui-he-dong-tai-gui-hua-si-lu-xiang-jie-by-jerry_n/

【总结】

1.动态规划流程

第 1 步：设计状态
第 2 步：状态转移方程
第 3 步：考虑初始化
第 4 步：考虑输出
第 5 步：考虑是否可以状态压缩

2.优先队列 按大小弹出 溢出问题long 去重set

参考链接：
https://leetcode-cn.com/problems/chou-shu-lcof/solution/dui-he-dong-tai-gui-hua-si-lu-xiang-jie-by-jerry_n/
参考链接：https://leetcode-cn.com/problems/chou-shu-lcof/solution/mian-shi-ti-49-chou-shu-dong-tai-gui-hua-qing-xi-t/